:: Abstract Reduction Systems and Idea of {K}nuth {B}endix Completion
:: Algorithm
:: by Grzegorz Bancerek
::
:: Received March 31, 2014
:: Copyright (c) 2014-2021 Association of Mizar Users
:: (Stowarzyszenie Uzytkownikow Mizara, Bialystok, Poland).
:: This code can be distributed under the GNU General Public Licence
:: version 3.0 or later, or the Creative Commons Attribution-ShareAlike
:: License version 3.0 or later, subject to the binding interpretation
:: detailed in file COPYING.interpretation.
:: See COPYING.GPL and COPYING.CC-BY-SA for the full text of these
:: licenses, or see http://www.gnu.org/licenses/gpl.html and
:: http://creativecommons.org/licenses/by-sa/3.0/.
environ
vocabularies RELAT_1, XBOOLE_0, FUNCT_1, REWRITE1, TDGROUP, ABSRED_0,
ZFMISC_1, FINSEQ_1, ARYTM_1, SUBSET_1, NUMBERS, STRUCT_0, NAT_1, ARYTM_3,
CARD_1, XXREAL_0, ZFREFLE1, TARSKI, UNIALG_1, GROUP_1, MSUALG_6, FUNCT_2,
INCPROJ, EQREL_1, MSUALG_1, PARTFUN1, UNIALG_2, FUNCT_4, PBOOLE, FUNCT_7,
FINSEQ_2, FUNCOP_1, ORDINAL1, MESFUNC1;
notations TARSKI, XBOOLE_0, ZFMISC_1, ENUMSET1, NUMBERS, XCMPLX_0, XXREAL_0,
RELAT_1, RELSET_1, FUNCT_1, SUBSET_1, PARTFUN1, FUNCT_2, FUNCOP_1,
EQREL_1, ORDINAL1, BINOP_1, FINSEQ_1, FINSEQ_2, NAT_1, FINSEQ_4, FUNCT_7,
MARGREL1, STRUCT_0, PBOOLE, UNIALG_1, PUA2MSS1, REWRITE1;
constructors RELAT_1, RELSET_1, FUNCT_2, STRUCT_0, REWRITE1, XCMPLX_0,
XXREAL_0, NAT_1, FINSEQ_5, ENUMSET1, BINOP_1, FINSEQ_1, FINSEQ_4,
FUNCT_7, CARD_1, XXREAL_1, UNIALG_1, PUA2MSS1, REALSET1, MARGREL1,
EQREL_1, NUMBERS, XBOOLE_0, ZFMISC_1, SUBSET_1, FUNCT_1, PARTFUN1;
registrations SUBSET_1, XBOOLE_0, RELSET_1, ORDINAL1, NAT_1, REWRITE1,
XXREAL_0, XCMPLX_0, STRUCT_0, AOFA_A00, FUNCT_2, FINSEQ_1, PARTFUN1,
FUNCOP_1, FINSEQ_2, CARD_1, MARGREL1, UNIALG_1, PUA2MSS1, RELAT_1;
requirements BOOLE, SUBSET, NUMERALS, ARITHM, REAL;
definitions MARGREL1, STRUCT_0, UNIALG_1, REWRITE1;
equalities EQREL_1;
theorems ZFMISC_1, NAT_1, FINSEQ_3, FINSEQ_5, REWRITE1, IDEA_1, XBOOLE_0,
RELAT_1, FUNCT_1, FUNCT_2, TARSKI, SUBSET_1, ENUMSET1, SETWISEO,
ORDINAL1, SEQ_4, MARGREL1, RELSET_1, FINSEQ_1, PARTFUN1, FINSEQ_2,
GRFUNC_1, FUNCOP_1, FUNCT_7, PUA2MSS1, COMPUT_1;
schemes NAT_1, RECDEF_1, RELSET_1;
begin :: Reduction and Convertibility
definition
struct(1-sorted) ARS(#
carrier -> set,
reduction -> Relation of the carrier
#);
end;
registration
let A be non empty set, r be Relation of A;
cluster ARS(#A, r#) -> non empty;
coherence;
end;
registration
cluster non empty strict for ARS;
existence
proof
set A = the non empty set, r = the Relation of A;
take X = ARS(#A,r#);
thus X is non empty;
thus X is strict;
end;
end;
definition
let X be ARS;
let x,y be Element of X;
pred x ==> y means [x,y] in the reduction of X;
end;
notation
let X be ARS;
let x,y be Element of X;
synonym y <== x for x ==> y;
end;
definition
let X be ARS;
let x,y be Element of X;
pred x =01=> y means x = y or x ==> y;
reflexivity;
pred x =*=> y means the reduction of X reduces x,y;
reflexivity by REWRITE1:12;
end;
reserve X for ARS, a,b,c,u,v,w,x,y,z for Element of X;
theorem
a ==> b implies X is non empty;
theorem Th2:
x ==> y implies x =*=> y by REWRITE1:15;
theorem Th3:
x =*=> y & y =*=> z implies x =*=> z by REWRITE1:16;
scheme Star{X() -> ARS, P[object]}:
for x,y being Element of X() st x =*=> y & P[x]
holds P[y]
provided
A1: for x,y being Element of X() st x ==> y & P[x] holds P[y]
proof
let x,y be Element of X();
given p being RedSequence of the reduction of X() such that
A2: p.1 = x & p.len p = y;
assume
A0: P[x];
defpred Q[Nat] means $1+1 in dom p implies P[p.($1+1)];
A3: Q[0] by A0,A2;
A4: for i being Nat st Q[i] holds Q[i+1]
proof
let i be Nat; reconsider j = i as Element of NAT by ORDINAL1:def 12;
assume
B1: Q[i] & i+1+1 in dom p; then
i+1+1 <= len p & i+1 >= 1 by NAT_1:11,FINSEQ_3:25; then
B2: j+1 in dom p by SEQ_4:134; then
[p.(i+1), p.(i+1+1)] in the reduction of X() by B1,REWRITE1:def 2; then
reconsider a = p.(i+1), b = p.(i+1+1) as Element of X() by ZFMISC_1:87;
P[a] & a ==> b by B1,B2,REWRITE1:def 2;
hence P[p.(i+1+1)] by A1;
end;
A5: for i being Nat holds Q[i] from NAT_1:sch 2(A3,A4);
len p >= 0+1 by NAT_1:13; then
(ex i being Nat st len p = 1+i) &
len p in dom p by NAT_1:10,FINSEQ_5:6;
hence thesis by A2,A5;
end;
scheme Star1{X() -> ARS, P[object], a, b() -> Element of X()}:
P[b()]
provided
A1: a() =*=> b() and
A2: P[a()] and
A3: for x,y being Element of X() st x ==> y & P[x] holds P[y]
proof
for x,y being Element of X() st x =*=> y & P[x] holds P[y] from Star(A3);
hence thesis by A1,A2;
end;
scheme StarBack{X() -> ARS, P[object]}:
for x,y being Element of X() st x =*=> y & P[y]
holds P[x]
provided
A1: for x,y being Element of X() st x ==> y & P[y] holds P[x]
proof
let x,y be Element of X();
given p being RedSequence of the reduction of X() such that
A2: p.1 = x & p.len p = y;
assume
A0: P[y];
defpred Q[Nat] means (len p)-$1 in dom p implies P[p.((len p)-$1)];
A3: Q[0] by A0,A2;
A4: for i being Nat st Q[i] holds Q[i+1]
proof
let i be Nat; assume
B1: Q[i] & (len p)-(i+1) in dom p; then
reconsider k = (len p)-(i+1) as Element of NAT;
B4: k >= 0+1 by B1,FINSEQ_3:25;
i is Element of NAT &
k+1 = (len p)-i by ORDINAL1:def 12; then
k+1 <= len p by IDEA_1:3; then
B2: k in dom p & k+1 in dom p by B4,SEQ_4:134; then
[p.k, p.(k+1)] in the reduction of X() by REWRITE1:def 2; then
reconsider a = p.k, b = p.(k+1) as Element of X() by ZFMISC_1:87;
P[b] & a ==> b by B1,B2,REWRITE1:def 2;
hence thesis by A1;
end;
A5: for i being Nat holds Q[i] from NAT_1:sch 2(A3,A4);
len p >= 0+1 by NAT_1:13; then
len p-1 is Nat & (len p)-((len p)-1) = 1 & 1 in dom p
by NAT_1:21,FINSEQ_5:6;
hence thesis by A2,A5;
end;
scheme StarBack1{X() -> ARS, P[object], a, b() -> Element of X()}:
P[a()]
provided
A1: a() =*=> b() and
A2: P[b()] and
A3: for x,y being Element of X() st x ==> y & P[y] holds P[x]
proof
for x,y being Element of X() st x =*=> y & P[y] holds P[x]
from StarBack(A3);
hence thesis by A1,A2;
end;
definition
let X be ARS;
let x,y be Element of X;
pred x =+=> y means ex z being Element of X st x ==> z & z =*=> y;
end;
theorem Th4:
x =+=> y iff ex z st x =*=> z & z ==> y
proof
thus x =+=> y implies ex z st x =*=> z & z ==> y
proof given z such that
A1: x ==> z & z =*=> y;
defpred P[Element of X] means ex u st x =*=> u & u ==> $1;
A2: for y,z st y ==> z & P[y] holds P[z]
proof
let y,z; assume
A3: y ==> z;
given u such that
A4: x =*=> u & u ==> y;
take y;
u =*=> y by A4,Th2;
hence thesis by A3,A4,Th3;
end;
A5: for y,z st y =*=> z & P[y] holds P[z] from Star(A2);
thus thesis by A1,A5;
end;
given z such that
A6: x =*=> z & z ==> y;
defpred P[Element of X] means ex u st $1 ==> u & u =*=> y;
A2: for y,z st y ==> z & P[z] holds P[y]
proof
let x,z; assume
A3: x ==> z;
given u such that
A4: z ==> u & u =*=> y;
take z;
z =*=> u by A4,Th2;
hence thesis by A3,A4,Th3;
end;
A5: for y,z st y =*=> z & P[z] holds P[y] from StarBack(A2);
thus ex z st x ==> z & z =*=> y by A5,A6;
end;
notation
let X,x,y;
synonym y <=01= x for x =01=> y;
synonym y <=*= x for x =*=> y;
synonym y <=+= x for x =+=> y;
end;
:: x ==> y implies x =+=> y;
:: x =+=> y implies x =*=> y;
:: x =+=> y & y =*=> z implies x =+=> z;
:: x =*=> y & y =+=> z implies x =+=> z;
definition
let X,x,y;
pred x <==> y means x ==> y or x <== y;
symmetry;
end;
theorem
x <==> y iff [x,y] in (the reduction of X)\/(the reduction of X)~
proof
A1: x ==> y iff [x,y] in the reduction of X;
A2: x <== y iff [y,x] in the reduction of X;
[y,x] in the reduction of X iff [x,y] in (the reduction of X)~
by RELAT_1:def 7;
hence thesis by A1,A2,XBOOLE_0:def 3;
end;
definition
let X,x,y;
pred x <=01=> y means x = y or x <==> y;
reflexivity;
symmetry;
pred x <=*=> y means x,y are_convertible_wrt the reduction of X;
reflexivity by REWRITE1:26;
symmetry by REWRITE1:31;
end;
theorem Th6:
x <==> y implies x <=*=> y
proof
assume x ==> y or x <== y;
hence x,y are_convertible_wrt the reduction of X by REWRITE1:29,31;
end;
theorem Th7:
x <=*=> y & y <=*=> z implies x <=*=> z by REWRITE1:30;
scheme Star2{X() -> ARS, P[object]}:
for x,y being Element of X() st x <=*=> y & P[x]
holds P[y]
provided
A1: for x,y being Element of X() st x <==> y & P[x]
holds P[y]
proof
let x,y be Element of X();
set R = the reduction of X();
assume R\/R~ reduces x,y; then :: Only 2 expansions?
:: given p being RedSequence of R\/R~ such that
consider p being RedSequence of R\/R~ such that
A2: p.1 = x & p.len p = y by REWRITE1:def 3;
assume
A0: P[x];
defpred Q[Nat] means $1+1 in dom p implies P[p.($1+1)];
A3: Q[0] by A0,A2;
A4: for i being Nat st Q[i] holds Q[i+1]
proof
let i be Nat; reconsider j = i as Element of NAT by ORDINAL1:def 12;
assume
B1: Q[i] & i+1+1 in dom p; then
B4: i+1+1 <= len p & i+1 >= 1 by NAT_1:11,FINSEQ_3:25; then
j+1 in dom p by SEQ_4:134; then
B3: [p.(i+1), p.(i+1+1)] in R\/R~ by B1,REWRITE1:def 2; then
reconsider a = p.(i+1), b = p.(i+1+1) as Element of X() by ZFMISC_1:87;
[a,b] in R or [a,b] in R~ by B3,XBOOLE_0:def 3; then
a ==> b or b ==> a by RELAT_1:def 7; then
P[a] & a <==> b by B1,B4,SEQ_4:134;
hence P[p.(i+1+1)] by A1;
end;
A5: for i being Nat holds Q[i] from NAT_1:sch 2(A3,A4);
len p >= 0+1 by NAT_1:13; then
(ex i being Nat st len p = 1+i) &
len p in dom p by NAT_1:10,FINSEQ_5:6;
hence thesis by A2,A5;
end;
scheme Star2A{X() -> ARS, P[object], a, b() -> Element of X()}:
P[b()]
provided
A1: a() <=*=> b() and
A2: P[a()] and
A3: for x,y being Element of X() st x <==> y & P[x] holds P[y]
proof
for x,y being Element of X() st x <=*=> y & P[x] holds P[y] from Star2(A3);
hence thesis by A1,A2;
end;
definition
let X,x,y;
pred x <=+=> y means: Def8: ex z st x <==> z & z <=*=> y;
symmetry
proof let x,y;
given z such that
A1: x <==> z & z <=*=> y;
defpred P[Element of X] means ex u st x <=*=> u & u <==> $1;
A2: for y,z st y <==> z & P[y] holds P[z]
proof
let y,z; assume
A3: y <==> z;
given u such that
A4: x <=*=> u & u <==> y;
take y;
u <=*=> y by A4,Th6;
hence thesis by A3,A4,Th7;
end;
A5: for y,z st y <=*=> z & P[y] holds P[z] from Star2(A2);
ex u st x <=*=> u & u <==> y by A1,A5;
hence thesis;
end;
end;
theorem Th8:
x <=+=> y iff ex z st x <=*=> z & z <==> y
proof
x <=+=> y iff ex z st y <==> z & z <=*=> x by Def8;
hence thesis;
end;
theorem Lem1:
x =01=> y implies x =*=> y by Th2;
theorem Lem2:
x =+=> y implies x =*=> y
proof
assume
A1: x =+=> y;
consider z such that
A2: x ==> z & z =*=> y by A1;
A3: x =*=> z by A2,Th2;
thus x =*=> y by A2,A3,Th3;
end;
theorem
x ==> y implies x =+=> y;
theorem Lem3:
x ==> y & y ==> z implies x =*=> z
proof
assume
A1: x ==> y;
assume
A2: y ==> z;
A3: x =*=> y by A1,Th2;
A4: y =*=> z by A2,Th2;
thus x =*=> z by A3,A4,Th3;
end;
theorem Lem4:
x ==> y & y =01=> z implies x =*=> z
proof
assume
A1: x ==> y;
assume
A2: y =01=> z;
A3: x =*=> y by A1,Th2;
A4: y =*=> z by A2,Lem1;
thus x =*=> z by A3,A4,Th3;
end;
theorem Lem5:
x ==> y & y =*=> z implies x =*=> z
proof
assume
A1: x ==> y;
assume
A2: y =*=> z;
A3: x =*=> y by A1,Th2;
thus x =*=> z by A3,A2,Th3;
end;
theorem Lem5A:
x ==> y & y =+=> z implies x =*=> z
proof
assume
A1: x ==> y;
assume
A2: y =+=> z;
A3: x =*=> y by A1,Th2;
A4: y =*=> z by A2,Lem2;
thus x =*=> z by A3,A4,Th3;
end;
theorem
x =01=> y & y ==> z implies x =*=> z
proof
assume
A1: x =01=> y;
assume
A2: y ==> z;
A3: x =*=> y by A1,Lem1;
A4: y =*=> z by A2,Th2;
thus x =*=> z by A3,A4,Th3;
end;
theorem
x =01=> y & y =01=> z implies x =*=> z
proof
assume
A1: x =01=> y;
assume
A2: y =01=> z;
A3: x =*=> y by A1,Lem1;
A4: y =*=> z by A2,Lem1;
thus x =*=> z by A3,A4,Th3;
end;
theorem Lem8:
x =01=> y & y =*=> z implies x =*=> z
proof
assume
A1: x =01=> y;
assume
A2: y =*=> z;
A3: x =*=> y by A1,Lem1;
thus x =*=> z by A3,A2,Th3;
end;
theorem
x =01=> y & y =+=> z implies x =*=> z
proof
assume
A1: x =01=> y;
assume
A2: y =+=> z;
A3: x =*=> y by A1,Lem1;
A4: y =*=> z by A2,Lem2;
thus x =*=> z by A3,A4,Th3;
end;
theorem Lem10:
x =*=> y & y ==> z implies x =*=> z
proof
assume
A1: x =*=> y;
assume
A2: y ==> z;
A4: y =*=> z by A2,Th2;
thus x =*=> z by A1,A4,Th3;
end;
theorem Lem11:
x =*=> y & y =01=> z implies x =*=> z
proof
assume
A1: x =*=> y;
assume
A2: y =01=> z;
A4: y =*=> z by A2,Lem1;
thus x =*=> z by A1,A4,Th3;
end;
theorem Lem11A:
x =*=> y & y =+=> z implies x =*=> z
proof
assume
A1: x =*=> y;
assume
A2: y =+=> z;
A4: y =*=> z by A2,Lem2;
thus x =*=> z by A1,A4,Th3;
end;
theorem
x =+=> y & y ==> z implies x =*=> z
proof
assume
A1: x =+=> y;
assume
A2: y ==> z;
A3: x =*=> y by A1,Lem2;
A4: y =*=> z by A2,Th2;
thus x =*=> z by A3,A4,Th3;
end;
theorem
x =+=> y & y =01=> z implies x =*=> z
proof
assume
A1: x =+=> y;
assume
A2: y =01=> z;
A3: x =*=> y by A1,Lem2;
A4: y =*=> z by A2,Lem1;
thus x =*=> z by A3,A4,Th3;
end;
theorem
x =+=> y & y =+=> z implies x =*=> z
proof
assume
A1: x =+=> y;
assume
A2: y =+=> z;
A3: x =*=> y by A1,Lem2;
A4: y =*=> z by A2,Lem2;
thus x =*=> z by A3,A4,Th3;
end;
theorem
x ==> y & y ==> z implies x =+=> z by Th2;
theorem
x ==> y & y =01=> z implies x =+=> z by Lem1;
theorem
x ==> y & y =+=> z implies x =+=> z by Lem2;
theorem
x =01=> y & y ==> z implies x =+=> z by Lem1,Th4;
theorem
x =01=> y & y =+=> z implies x =+=> z
proof
assume
A1: x =01=> y;
assume
A2: y =+=> z;
consider u such that
A3: y =*=> u & u ==> z by A2,Th4;
thus x =+=> z by A3,A1,Lem8,Th4;
end;
theorem
x =*=> y & y =+=> z implies x =+=> z
proof
assume
A1: x =*=> y;
assume
A2: y =+=> z;
consider u such that
A3: y =*=> u & u ==> z by A2,Th4;
thus x =+=> z by A3,A1,Th3,Th4;
end;
theorem
x =+=> y & y ==> z implies x =+=> z by Lem10;
theorem
x =+=> y & y =01=> z implies x =+=> z by Lem11;
theorem
x =+=> y & y =*=> z implies x =+=> z by Th3;
theorem
x =+=> y & y =+=> z implies x =+=> z by Lem11A;
theorem Lem1A:
x <=01=> y implies x <=*=> y by Th6;
theorem Lem2A:
x <=+=> y implies x <=*=> y
proof
assume
A1: x <=+=> y;
consider z such that
A2: x <==> z & z <=*=> y by A1;
A3: x <=*=> z by A2,Th6;
thus x <=*=> y by A2,A3,Th7;
end;
theorem LemB:
x <==> y implies x <=+=> y;
theorem
x <==> y & y <==> z implies x <=*=> z
proof
assume
A1: x <==> y;
assume
A2: y <==> z;
A3: x <=*=> y by A1,Th6;
A4: y <=*=> z by A2,Th6;
thus x <=*=> z by A3,A4,Th7;
end;
theorem Lem4A:
x <==> y & y <=01=> z implies x <=*=> z
proof
assume
A1: x <==> y;
assume
A2: y <=01=> z;
A3: x <=*=> y by A1,Th6;
A4: y <=*=> z by A2,Lem1A;
thus x <=*=> z by A3,A4,Th7;
end;
theorem
x <=01=> y & y <==> z implies x <=*=> z by Lem4A;
theorem Lem5a:
x <==> y & y <=*=> z implies x <=*=> z
proof
assume
A1: x <==> y;
assume
A2: y <=*=> z;
A3: x <=*=> y by A1,Th6;
thus x <=*=> z by A3,A2,Th7;
end;
theorem
x <=*=> y & y <==> z implies x <=*=> z by Lem5a;
theorem Lem5B:
x <==> y & y <=+=> z implies x <=*=> z
proof
assume
A1: x <==> y;
assume
A2: y <=+=> z;
A3: x <=*=> y by A1,Th6;
A4: y <=*=> z by A2,Lem2A;
thus x <=*=> z by A3,A4,Th7;
end;
theorem
x <=+=> y & y <==> z implies x <=*=> z by Lem5B;
theorem
x <=01=> y & y <=01=> z implies x <=*=> z
proof
assume
A1: x <=01=> y;
assume
A2: y <=01=> z;
A3: x <=*=> y by A1,Lem1A;
A4: y <=*=> z by A2,Lem1A;
thus x <=*=> z by A3,A4,Th7;
end;
theorem Lm8:
x <=01=> y & y <=*=> z implies x <=*=> z
proof
assume
A1: x <=01=> y;
assume
A2: y <=*=> z;
A3: x <=*=> y by A1,Lem1A;
thus x <=*=> z by A3,A2,Th7;
end;
theorem
x <=*=> y & y <=01=> z implies x <=*=> z by Lm8;
theorem Lem9:
x <=01=> y & y <=+=> z implies x <=*=> z
proof
assume
A1: x <=01=> y;
assume
A2: y <=+=> z;
A3: x <=*=> y by A1,Lem1A;
A4: y <=*=> z by A2,Lem2A;
thus x <=*=> z by A3,A4,Th7;
end;
theorem
x <=+=> y & y <=01=> z implies x <=*=> z by Lem9;
theorem Lem11A:
x <=*=> y & y <=+=> z implies x <=*=> z
proof
assume
A1: x <=*=> y;
assume
A2: y <=+=> z;
A4: y <=*=> z by A2,Lem2A;
thus x <=*=> z by A1,A4,Th7;
end;
theorem
x <=+=> y & y <=+=> z implies x <=*=> z
proof
assume
A1: x <=+=> y;
assume
A2: y <=+=> z;
A3: x <=*=> y by A1,Lem2A;
A4: y <=*=> z by A2,Lem2A;
thus x <=*=> z by A3,A4,Th7;
end;
theorem
x <==> y & y <==> z implies x <=+=> z by Th6;
theorem
x <==> y & y <=01=> z implies x <=+=> z by Lem1A;
theorem
x <==> y & y <=+=> z implies x <=+=> z by Lem2A;
theorem Lem18:
x <=01=> y & y <=+=> z implies x <=+=> z
proof
assume
A1: x <=01=> y;
assume
A2: y <=+=> z;
consider u such that
A3: y <=*=> u & u <==> z by A2,Th8;
thus x <=+=> z by A3,A1,Lm8,Th8;
end;
theorem
x <=*=> y & y <=+=> z implies x <=+=> z
proof
assume
A1: x <=*=> y;
assume
A2: y <=+=> z;
consider u such that
A3: y <=*=> u & u <==> z by A2,Th8;
thus x <=+=> z by A3,A1,Th7,Th8;
end;
theorem
x <=+=> y & y <=+=> z implies x <=+=> z by Lem11A;
theorem Lem31:
x <=01=> y implies x <== y or x = y or x ==> y
proof
assume
A1: x <=01=> y;
A2: x <==> y or x = y by A1;
thus x <== y or x = y or x ==> y by A2;
end;
theorem
x <== y or x = y or x ==> y implies x <=01=> y
proof
assume
A1: x <== y or x = y or x ==> y;
A2: x <==> y or x = y by A1;
thus x <=01=> y by A2;
end;
theorem
x <=01=> y implies x <=01= y or x ==> y
proof
assume
A1: x <=01=> y;
A2: x <==> y or x = y by A1;
thus x <=01= y or x ==> y by A2;
end;
theorem
x <=01= y or x ==> y implies x <=01=> y
proof
assume
A1: x <=01= y or x ==> y;
A3: x <==> y or x = y by A1;
thus x <=01=> y by A3;
end;
theorem
x <=01=> y implies x <=01= y or x =+=> y
proof
assume
A1: x <=01=> y;
A2: x <==> y or x = y by A1;
thus x <=01= y or x =+=> y by A2;
end;
theorem
x <=01=> y implies x <=01= y or x <==> y;
theorem
x <=01= y or x <==> y implies x <=01=> y
proof
assume
A1: x <=01= y or x <==> y;
A3: x = y or x <==> y by A1;
thus x <=01=> y by A3;
end;
theorem
x <=*=> y & y ==> z implies x <=+=> z
proof
assume
A1: x <=*=> y;
assume
A2: y ==> z;
A4: y <==> z by A2;
thus x <=+=> z by A1,A4,Def8;
end;
theorem
x <=+=> y & y ==> z implies x <=+=> z
proof
assume
A1: x <=+=> y;
assume
A2: y ==> z;
A3: x <=*=> y by A1,Lem2A;
A4: y <==> z by A2;
thus x <=+=> z by A3,A4,Def8;
end;
theorem
x <=01=> y implies x <=01= y or x ==> y
proof
assume
A1: x <=01=> y;
A2: x = y or x <==> y by A1;
thus x <=01= y or x ==> y by A2;
end;
theorem
x <=01=> y implies x <=01= y or x =+=> y
proof
assume
A1: x <=01=> y;
A2: x = y or x <==> y by A1;
thus x <=01= y or x =+=> y by A2;
end;
theorem Lem43:
x <=01= y or x ==> y implies x <=01=> y
proof
assume
A1: x <=01= y or x ==> y;
A3: x <==> y or x = y by A1;
thus x <=01=> y by A3;
end;
theorem
x <=01= y or x <==> y implies x <=01=> y
proof
assume
A1: x <=01= y or x <==> y;
A3: x <==> y or x = y by A1;
thus x <=01=> y by A3;
end;
theorem
x <=01=> y implies x <=01= y or x <==> y;
theorem
x <=+=> y & y ==> z implies x <=+=> z
proof
assume
A1: x <=+=> y;
assume
A2: y ==> z;
A3: x <=*=> y by A1,Lem2A;
A4: y <==> z by A2;
thus x <=+=> z by A3,A4,Def8;
end;
theorem
x <=*=> y & y ==> z implies x <=+=> z
proof
assume
A1: x <=*=> y;
assume
A2: y ==> z;
A4: y <==> z by A2;
thus x <=+=> z by A1,A4,Def8;
end;
theorem
x <=01=> y & y ==> z implies x <=+=> z
proof
assume
A1: x <=01=> y;
assume
A2: y ==> z;
A4: y <==> z by A2;
thus x <=+=> z by A1,A4,Lem1A,Def8;
end;
theorem
x <=+=> y & y =01=> z implies x <=+=> z
proof
assume
A1: x <=+=> y;
assume
A2: y =01=> z;
A3: y <=01=> z by A2,Lem43;
thus x <=+=> z by A1,A3,Lem18;
end;
theorem
x <==> y & y =01=> z implies x <=+=> z
proof
assume
A1: x <==> y;
assume
A2: y =01=> z;
A3: y <=01=> z by A2,Lem43;
thus x <=+=> z by A3,A1,LemB,Lem18;
end;
theorem
x ==> y & y ==> z & z ==> u implies x =+=> u by Lem3;
theorem
x ==> y & y =01=> z & z ==> u implies x =+=> u by Lem4,Th4;
theorem
x ==> y & y =*=> z & z ==> u implies x =+=> u by Lem5,Th4;
theorem
x ==> y & y =+=> z & z ==> u implies x =+=> u
proof
assume
A1: x ==> y;
assume
A2: y =+=> z;
assume
A3: z ==> u;
A4: x =*=> z by A1,A2,Lem5A;
thus x =+=> u by A3,A4,Th4;
end;
theorem LemZ:
x =*=> y implies x <=*=> y
proof
assume
A1: x =*=> y;
defpred P[Element of X] means x <=*=> $1;
A2: P[x];
A3: for y,z st y ==> z & P[y] holds P[z]
proof
let y,z;
assume
A4: y ==> z;
assume
A5: P[y];
A6: y <==> z by A4;
A7: y <=*=> z by A6,Th6;
thus P[z] by A5,A7,Th7;
end;
thus P[y] from Star1(A1,A2,A3);
end;
theorem
for z st
for x,y st x ==> z & x ==> y holds y ==> z
for x,y st x ==> z & x =*=> y
holds y ==> z
proof
let z;
assume
A: for x,y st x ==> z & x ==> y holds y ==> z;
let x,y;
assume
B: x ==> z & x =*=> y;
defpred P[Element of X] means $1 ==> z;
C: for u,v st u ==> v & P[u] holds P[v] by A;
D: for u,v st u =*=> v & P[u] holds P[v] from Star(C);
thus y ==> z by B,D;
end;
theorem
(for x,y st x ==> y holds y ==> x)
implies
for x,y st x <=*=> y holds x =*=> y
proof
assume
A: for x,y st x ==> y holds y ==> x;
let x,y;
assume
B: x <=*=> y;
defpred P[Element of X] means x =*=> $1;
C: for u,v st u <==> v & P[u] holds P[v] by A,Lem10;
D: for u,v st u <=*=> v & P[u] holds P[v] from Star2(C);
thus x =*=> y by B,D;
end;
theorem LemN:
x =*=> y implies x = y or x =+=> y
proof
assume
A1: x =*=> y;
defpred P[Element of X] means x = $1 or x =+=> $1;
A2: P[x];
A3: for y,z st y ==> z & P[y] holds P[z]
proof
let y,z;
assume
A4: y ==> z;
assume
A5: P[y];
A6: x =*=> y by A5,Lem2;
thus P[z] by A6,A4,Th4;
end;
thus P[y] from Star1(A1,A2,A3);
end;
theorem
(for x,y,z st x ==> y & y ==> z holds x ==> z)
implies
for x,y st x =+=> y holds x ==> y
proof
assume
A1: for x,y,z st x ==> y & y ==> z holds x ==> z;
let x,y;
assume
A2: x =+=> y;
consider z such that
A3: x ==> z and
A4: z =*=> y by A2;
defpred P[Element of X] means x ==> $1;
A5: P[z] by A3;
A6: for u,v st u ==> v & P[u] holds P[v] by A1;
thus P[y] from Star1(A4,A5,A6);
end;
begin :: Examples of ARS
scheme ARSex{A() -> non empty set, R[object,object]}:
ex X being strict non empty ARS st the carrier of X = A() &
for x,y being Element of X holds x ==> y iff R[x,y]
proof
consider r being Relation of A() such that
A1: for x,y being Element of A() holds [x,y] in r iff R[x,y]
from RELSET_1:sch 2;
take X = ARS(#A(), r#);
thus the carrier of X = A();
thus thesis by A1;
end;
definition
func ARS_01 -> strict ARS means:
Def18:
the carrier of it = {0,1} &
the reduction of it = [:{0},{0,1}:];
existence
proof
{0} c= {0,1} by ZFMISC_1:7; then
reconsider r = [:{0},{0,1}:] as Relation of {0,1} by ZFMISC_1:96;
take X = ARS(#{0,1}, r#);
thus thesis;
end;
uniqueness;
func ARS_02 -> strict ARS means:
Def19:
the carrier of it = {0,1,2} &
the reduction of it = [:{0},{0,1,2}:];
existence
proof
{0} c= {0,1,2} by SETWISEO:1; then
reconsider r = [:{0},{0,1,2}:] as Relation of {0,1,2} by ZFMISC_1:96;
take X = ARS(#{0,1,2}, r#);
thus thesis;
end;
uniqueness;
end;
registration
cluster ARS_01 -> non empty;
coherence by Def18;
cluster ARS_02 -> non empty;
coherence by Def19;
end;
reserve i,j,k for Element of ARS_01;
theorem ThA1:
for s being set holds s is Element of ARS_01 iff s = 0 or s = 1
proof
let s be set;
the carrier of ARS_01 = {0,1} by Def18;
hence thesis by TARSKI:def 2;
end;
theorem
i ==> j iff i = 0
proof
the reduction of ARS_01 = [:{0},{0,1}:] by Def18; then
i ==> j iff i in {0} & j in {0,1} by ZFMISC_1:87; then
i ==> j iff i = 0 & (j = 0 or j = 1) by TARSKI:def 1,def 2;
hence thesis by ThA1;
end;
reserve l,m,n for Element of ARS_02;
theorem ThB1:
for s being set holds s is Element of ARS_02 iff s = 0 or s = 1 or s = 2
proof
let s be set;
the carrier of ARS_02 = {0,1,2} by Def19;
hence thesis by ENUMSET1:def 1;
end;
theorem
m ==> n iff m = 0
proof
the reduction of ARS_02 = [:{0},{0,1,2}:] by Def19; then
m ==> n iff m in {0} & n in {0,1,2} by ZFMISC_1:87; then
m ==> n iff m = 0 & (n = 0 or n = 1 or n = 2)
by TARSKI:def 1,ENUMSET1:def 1;
hence thesis by ThB1;
end;
begin :: Normal Forms
definition
let X,x;
attr x is normform means not ex y st x ==> y;
end;
theorem Ch1:
x is normform iff x is_a_normal_form_wrt the reduction of X
proof set R = the reduction of X;
thus x is normform implies x is_a_normal_form_wrt the reduction of X
proof assume
Z0: not ex y st x ==> y;
let a be object;
assume
Z1: [x,a] in the reduction of X; then
reconsider y = a as Element of X by ZFMISC_1:87;
x ==> y by Z1;
hence thesis by Z0;
end;
assume
Z1: not ex b being object st [x,b] in R;
let y;
assume [x,y] in the reduction of X;
hence thesis by Z1;
end;
definition
let X,x,y;
pred x is_normform_of y means x is normform & y =*=> x;
end;
theorem Ch2:
x is_normform_of y iff x is_a_normal_form_of y, the reduction of X
proof set R = the reduction of X;
thus x is_normform_of y implies x is_a_normal_form_of y, R
proof assume
x is normform & R reduces y,x;
hence x is_a_normal_form_wrt R & R reduces y,x by Ch1;
end;
assume x is_a_normal_form_wrt R & R reduces y,x;
hence x is normform & R reduces y,x by Ch1;
end;
definition
let X,x;
attr x is normalizable means ex y st y is_normform_of x;
end;
theorem Ch3:
x is normalizable iff x has_a_normal_form_wrt the reduction of X
proof
set R = the reduction of X;
A0: field R c= (the carrier of X)\/the carrier of X by RELSET_1:8;
thus x is normalizable implies x has_a_normal_form_wrt R
proof
given y such that
A1: y is_normform_of x;
take y; thus thesis by A1,Ch2;
end;
given a being object such that
A2: a is_a_normal_form_of x, R;
R reduces x,a by A2,REWRITE1:def 6; then
x = a or a in field R by REWRITE1:18; then
reconsider a as Element of X by A0;
take a; thus thesis by A2,Ch2;
end;
definition
let X;
attr X is WN means for x holds x is normalizable;
attr X is SN means
for f being Function of NAT, the carrier of X
ex i being Nat st not f.i ==> f.(i+1);
attr X is UN* means
for x,y,z st y is_normform_of x & z is_normform_of x holds y = z;
attr X is UN means
for x,y st x is normform & y is normform & x <=*=> y holds x = y;
attr X is N.F. means
for x,y st x is normform & x <=*=> y holds y =*=> x;
end;
theorem
X is WN iff the reduction of X is weakly-normalizing
proof set R = the reduction of X;
A0: field R c= (the carrier of X)\/the carrier of X by RELSET_1:8;
thus X is WN implies R is weakly-normalizing
proof assume
A1: for x holds x is normalizable;
let a be object; assume a in field R; then
reconsider a as Element of X by A0;
a is normalizable by A1;
hence thesis by Ch3;
end;
assume
A2: for a being object st a in field R
holds a has_a_normal_form_wrt R;
let x;
per cases;
suppose
x in field R;
hence thesis by A2,Ch3;
end;
suppose
A3: not x in field R;
take x;
thus x is normform
proof
let y;
thus not [x,y] in R by A3,RELAT_1:15;
end;
thus thesis;
end;
end;
theorem Ch7:
X is SN implies the reduction of X is strongly-normalizing
proof set R = the reduction of X;
set A = the carrier of X;
A0: field R c= A \/ A by RELSET_1:8;
assume
A1: for f being Function of NAT, A
ex i being Nat st not f.i ==> f.(i+1);
let f be ManySortedSet of NAT;
per cases;
suppose f is A-valued; then
rng f c= A & dom f = NAT by RELAT_1:def 19,PARTFUN1:def 2; then
reconsider g = f as Function of NAT, A by FUNCT_2:2;
consider i being Nat such that
A2: not g.i ==> g.(i+1) by A1;
take i;
thus not [f.i,f.(i+1)] in R by A2;
end;
suppose
f is not A-valued; then
consider a being object such that
A3: a in rng f & not a in A by TARSKI:def 3,RELAT_1:def 19;
consider i being object such that
A4: i in dom f & a = f.i by A3,FUNCT_1:def 3;
reconsider i as Element of NAT by A4;
take i;
assume [f.i,f.(i+1)] in R; then
a in field R by A4,RELAT_1:15;
hence thesis by A0,A3;
end;
end;
theorem Ch8:
X is non empty & the reduction of X is strongly-normalizing implies X is SN
proof set R = the reduction of X;
set A = the carrier of X;
assume
A1: X is non empty;
assume
A5: for f being ManySortedSet of NAT
ex i being Nat st not [f.i,f.(i+1)] in R;
let f be Function of NAT, A;
consider i being Nat such that
A6: not [f.i,f.(i+1)] in R by A1,A5;
take i;
thus not [f.i,f.(i+1)] in R by A6;
end;
reserve A for set;
theorem ThSN:
for X holds X is SN iff
not ex A,z st z in A & for x st x in A ex y st y in A & x ==> y
proof
let X;
thus X is SN implies not ex A,z st z in A &
for x st x in A ex y st y in A & x ==> y
proof assume
00: for f being Function of NAT, the carrier of X
ex i being Nat st not f.i ==> f.(i+1);
given A,z such that
01: z in A & for x st x in A ex y st y in A & x ==> y;
ex y st y in A & z ==> y by 01; then
reconsider X0 = X as non empty ARS;
reconsider z0 = z as Element of X0;
defpred P[Nat,Element of X0,Element of X0] means
$2 in A implies $3 in A & $2 ==> $3;
02: for i being Nat, x being Element of X0
ex y being Element of X0 st P[i,x,y] by 01;
consider f being Function of NAT, the carrier of X0 such that
03: f.0 = z0 and
04: for i being Nat holds P[i,f.i,f.(i+1)]
from RECDEF_1:sch 2(02);
defpred Q[Nat] means f.$1 ==> f.($1+1) & f.$1 in A;
05: Q[0] by 01,03,04;
06: now let i be Nat; assume Q[i]; then
f.(i+1) in A by 04;
hence Q[i+1] by 04;
end;
for i being Nat holds Q[i] from NAT_1:sch 2(05,06);
hence contradiction by 00;
end;
assume
00: not ex A,z st z in A &
for x st x in A ex y st y in A & x ==> y;
given f being Function of NAT, the carrier of X such that
01: for i being Nat holds f.i ==> f.(i+1);
f.0 ==> f.(0+1) by 01; then
04: X is non empty & 0 in NAT by ORDINAL1:def 12; then
02: f.0 in rng f by FUNCT_2:4;
now let x; assume x in rng f; then
consider i being object such that
03: i in dom f & x = f.i by FUNCT_1:def 3;
reconsider i as Element of NAT by 03;
take y = f.(i+1);
thus y in rng f by 04,FUNCT_2:4;
thus x ==> y by 01,03;
end;
hence contradiction by 00,02;
end;
scheme notSN{X() -> ARS, P[object]}:
X() is not SN
provided
A1: ex x being Element of X() st P[x]
and
A2: for x being Element of X() st P[x]
ex y being Element of X() st P[y] & x ==> y
proof
set A = {x where x is Element of X(): P[x]};
consider z being Element of X() such that
A3: P[z] by A1;
A4: z in A by A3;
now let x be Element of X(); assume x in A; then
ex a being Element of X() st x = a & P[a]; then
consider y being Element of X() such that
A6: P[y] & x ==> y by A2;
take y; thus y in A by A6;
thus x ==> y by A6;
end;
hence thesis by A4,ThSN;
end;
theorem
X is UN iff the reduction of X is with_UN_property
proof
set R = the reduction of X;
set A = the carrier of X;
A0: field R c= A \/ A by RELSET_1:8;
thus X is UN implies R is with_UN_property
proof
assume
A1: for x,y st x is normform & y is normform & x <=*=> y holds x = y;
let a,b be object;
assume
A2: a is_a_normal_form_wrt R & b is_a_normal_form_wrt R &
a,b are_convertible_wrt R;
per cases;
suppose a in A & b in A; then
reconsider x = a, y = b as Element of X;
x is normform & y is normform & x <=*=> y by A2,Ch1;
hence a = b by A1;
end;
suppose not a in A or not b in A; then
not a in field R or not b in field R by A0;
hence a = b by A2,REWRITE1:28,31;
end;
end;
assume
A4: for a,b being object
st a is_a_normal_form_wrt R & b is_a_normal_form_wrt R &
a,b are_convertible_wrt R holds a = b;
let x,y; assume
x is normform & y is normform & x <=*=> y; then
x is_a_normal_form_wrt R & y is_a_normal_form_wrt R &
x,y are_convertible_wrt R by Ch1;
hence x = y by A4;
end;
theorem
X is N.F. iff the reduction of X is with_NF_property
proof
set R = the reduction of X;
set A = the carrier of X;
A0: field R c= A \/ A by RELSET_1:8;
thus X is N.F. implies R is with_NF_property
proof
assume
A1: for x,y st x is normform & x <=*=> y holds y =*=> x;
let a,b be object; assume
A2: a is_a_normal_form_wrt R & a,b are_convertible_wrt R;
per cases;
suppose a in A & b in A; then
reconsider x = a, y = b as Element of X;
x is normform & x <=*=> y by A2,Ch1; then
y =*=> x by A1;
hence R reduces b,a;
end;
suppose not a in A or not b in A; then
not a in field R or not b in field R by A0; then
a = b by A2,REWRITE1:28,31;
hence R reduces b,a by REWRITE1:12;
end;
end;
assume
B1: for a,b being object st
a is_a_normal_form_wrt R & a,b are_convertible_wrt R
holds R reduces b,a;
let x,y; assume
x is normform & x <=*=> y;
hence R reduces y,x by B1,Ch1;
end;
definition
let X;
let x such that
A: ex y st y is_normform_of x and
B: for y,z st y is_normform_of x & z is_normform_of x holds y = z;
func nf x -> Element of X means:
Def17:
it is_normform_of x;
existence by A;
uniqueness by B;
end;
theorem
(ex y st y is_normform_of x) &
(for y,z st y is_normform_of x & z is_normform_of x holds y = z)
implies nf x = nf(x, the reduction of X)
proof set R = the reduction of X;
set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
given y such that
A0: y is_normform_of x;
B0: x has_a_normal_form_wrt R by A0,Ch2,REWRITE1:def 11;
assume
A1: for y,z st y is_normform_of x & z is_normform_of x holds y = z; then
nf x is_normform_of x by A0,Def17; then
A2: nf x is_a_normal_form_of x,R by Ch2;
now
let b,c be object; assume
A3: b is_a_normal_form_of x,R & c is_a_normal_form_of x,R; then
A4: R reduces x,b & R reduces x,c by REWRITE1:def 6;
per cases;
suppose x in field R; then
b in field R & c in field R by A4,REWRITE1:19; then
reconsider y = b, z = c as Element of X by F0;
y is_normform_of x & z is_normform_of x by A3,Ch2;
hence b = c by A1;
end;
suppose not x in field R; then
x = b & x = c by A4,REWRITE1:18;
hence b = c;
end;
end;
hence nf x = nf(x, the reduction of X) by B0,A2,REWRITE1:def 12;
end;
theorem LemN1:
x is normform & x =*=> y implies x = y
proof
assume
A1: x is normform;
assume
A2: x =*=> y;
A4: not x =+=> y by A1;
thus x = y by A2,A4,LemN;
end;
theorem LemN2:
x is normform implies x is_normform_of x;
theorem
x is normform & y ==> x implies x is_normform_of y by Th2;
theorem
x is normform & y =01=> x implies x is_normform_of y by Lem1;
theorem
x is normform & y =+=> x implies x is_normform_of y by Lem2;
theorem
x is_normform_of y & y is_normform_of x implies x = y by LemN1;
theorem LemN6:
x is_normform_of y & z ==> y implies x is_normform_of z by Lem5;
theorem LemN7:
x is_normform_of y & z =*=> y implies x is_normform_of z by Th3;
theorem
x is_normform_of y & z =*=> x implies x is_normform_of z;
registration
let X;
cluster normform -> normalizable for Element of X;
coherence
proof let x;
assume
A1: x is normform;
take x;
thus x is_normform_of x by A1;
end;
end;
theorem LemN5:
x is normalizable & y ==> x implies y is normalizable by LemN6;
theorem ThWN1:
X is WN iff for x ex y st y is_normform_of x
proof
thus X is WN implies for x ex y st y is_normform_of x
proof
assume
A1: for x holds x is normalizable;
let x;
A2: x is normalizable by A1;
thus ex y st y is_normform_of x by A2;
end;
assume
A3: for x ex y st y is_normform_of x;
let x; thus ex y st y is_normform_of x by A3;
end;
theorem
(for x holds x is normform) implies X is WN
proof
assume
A1: for x holds x is normform;
let x;
A2: x is normform by A1;
thus ex y st y is_normform_of x by A2,LemN2;
end;
registration
cluster SN -> WN for ARS;
coherence
proof let X;
assume
A1: X is SN;
assume
A2: X is not WN;
consider z such that
A3: z is not normalizable by A2;
set A = {x: x is not normalizable};
A4: z in A by A3;
A5: for x st x in A ex y st y in A & x ==> y
proof
let x;
assume x in A; then
A6: ex y st x = y & y is not normalizable; then
x is not normform; then
consider y such that
A7: x ==> y;
take y;
y is not normalizable by A6,A7,LemN5;
hence y in A;
thus x ==> y by A7;
end;
thus contradiction by A1,A4,A5,ThSN;
end;
end;
theorem LmA:
x <> y & (for a,b holds a ==> b iff a = x)
implies y is normform & x is normalizable
proof
assume
Z0: x <> y;
assume
Z2: for a,b holds a ==> b iff a = x;
thus y is normform by Z0,Z2;
take y; thus y is normform by Z0,Z2;
thus thesis by Z2,Th2;
end;
theorem
ex X st X is WN & X is not SN
proof
defpred R[set,set] means $1 = 0;
consider X being strict non empty ARS such that
A1: the carrier of X = {0,1} and
A2: for x,y being Element of X holds x ==> y iff R[x,y] from ARSex;
reconsider z = 0, o = 1 as Element of X by A1,TARSKI:def 2;
A3: z <> o;
take X;
thus X is WN
proof
let x be Element of X;
x = 0 or x = 1 by A1,TARSKI:def 2; then
x is normform or x is normalizable by A2,A3,LmA;
hence thesis;
end;
set A = {z};
A4: z in A by TARSKI:def 1;
now
let x be Element of X;
assume x in A; then
A5: x = z by TARSKI:def 1;
take y = z;
thus y in A & x ==> y by A2,A5,TARSKI:def 1;
end;
hence X is not SN by A4,ThSN;
end;
registration
cluster N.F. -> UN* for ARS;
coherence
proof let X;
assume
A1: for x,y st x is normform & x <=*=> y holds y =*=> x;
let x,y,z;
assume
A2: y is normform & x =*=> y;
assume
A3: z is normform & x =*=> z;
A4: x <=*=> y & x <=*=> z by A2,A3,LemZ;
A5: y <=*=> z by A4,Th7;
thus y = z by A2,A1,A3,A5,LemN1;
end;
cluster N.F. -> UN for ARS;
coherence by LemN1;
cluster UN -> UN* for ARS;
coherence
proof let X;
assume
A1: for x,y st x is normform & y is normform & x <=*=> y holds x = y;
let x,y,z;
assume
A2: y is normform & x =*=> y;
assume
A3: z is normform & x =*=> z;
A4: x <=*=> y & x <=*=> z by A2,A3,LemZ;
thus y = z by A1,A2,A3,A4,Th7;
end;
end;
theorem LemN12:
X is WN UN* & x is normform & x <=*=> y implies y =*=> x
proof
assume
A1: X is WN UN*;
assume
A2: x is normform;
assume
A3: x <=*=> y;
defpred P[Element of X] means $1 =*=> x;
A4: for y,z st y <==> z & P[y] holds P[z]
proof
let y,z;
assume
B1: y <==> z;
assume
B2: P[y];
per cases by B1;
suppose
C1: y ==> z;
B3: z is normalizable by A1;
consider u such that
B4: u is_normform_of z by B3;
B5: u is_normform_of y by C1,B4,LemN6;
B6: x is_normform_of y by A2,B2;
thus P[z] by B4,B6,B5,A1;
end;
suppose
C2: y <== z;
thus P[z] by B2,C2,Lem5;
end;
end;
A5: for y,z st y <=*=> z & P[y] holds P[z] from Star2(A4);
thus y =*=> x by A3,A5;
end;
registration
cluster WN UN* -> N.F. for ARS;
coherence by LemN12;
cluster WN UN* -> UN for ARS;
coherence;
end;
theorem Lem21:
y is_normform_of x & z is_normform_of x & y <> z implies x =+=> y
proof
assume
A1: y is_normform_of x;
assume
A2: z is_normform_of x;
assume
A3: y <> z;
A6: x = y or x =+=> y by A1,LemN;
thus x =+=> y by A3,A1,A2,A6,LemN1;
end;
theorem Lem22:
X is WN UN* implies nf x is_normform_of x
proof
assume
A1: X is WN UN*;
A4: x is normalizable by A1;
A3: y is_normform_of x & z is_normform_of x implies y = z by A1;
thus nf x is_normform_of x by A4,A3,Def17;
end;
theorem Lem23:
X is WN UN* & y is_normform_of x implies y = nf x
proof
assume
A1: X is WN UN*;
assume
A2: y is_normform_of x;
A4: for z,u holds z is_normform_of x & u is_normform_of x implies z = u
by A1;
thus y = nf x by A2,A4,Def17;
end;
theorem Lem24:
X is WN UN* implies nf x is normform
proof
assume
A1: X is WN UN*;
A2: nf x is_normform_of x by A1,Lem22;
thus nf x is normform by A2;
end;
theorem
X is WN UN* implies nf nf x = nf x
proof
assume
A1: X is WN UN*;
A2: nf x is normform by A1,Lem24;
thus nf nf x = nf x by A1,A2,LemN2,Lem23;
end;
theorem Lem26:
X is WN UN* & x =*=> y implies nf x = nf y
proof
assume
A1: X is WN UN*;
assume
A2: x =*=> y;
A4: nf y is_normform_of x by A2,A1,Lem22,LemN7;
thus nf x = nf y by A1,A4,Lem23;
end;
theorem Lem27:
X is WN UN* & x <=*=> y implies nf x = nf y
proof
assume
A1: X is WN UN*;
assume
A2: x <=*=> y;
defpred P[Element of X] means nf x = nf $1;
A3: P[x];
A4: for z,u st z <==> u & P[z] holds P[u] by A1,Th2,Lem26;
P[y] from Star2A(A2,A3,A4);
hence thesis;
end;
theorem
X is WN UN* & nf x = nf y implies x <=*=> y
proof
assume
A1: X is WN UN*;
assume
A2: nf x = nf y;
nf x is_normform_of x & nf x is_normform_of y by A1,A2,Lem22; then
x <=*=> nf x & nf x <=*=> y by LemZ;
hence thesis by Th7;
end;
begin :: Divergence and Convergence
definition
let X,x,y;
pred x <<>> y means
ex z st x <=*= z & z =*=> y;
symmetry;
reflexivity;
pred x >><< y means:DEF2:
ex z st x =*=> z & z <=*= y;
symmetry;
reflexivity;
pred x <<01>> y means
ex z st x <=01= z & z =01=> y;
symmetry;
reflexivity;
pred x >>01<< y means
ex z st x =01=> z & z <=01= y;
symmetry;
reflexivity;
end;
theorem Ch11:
x <<>> y iff x,y are_divergent_wrt the reduction of X
proof set R = the reduction of X;
thus x <<>> y implies x,y are_divergent_wrt R
proof
given z such that
A1: x <=*= z & z =*=> y;
take z;
thus R reduces z,x & R reduces z,y by A1;
end;
set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
given a being object such that
A2: R reduces a,x & R reduces a,y;
per cases;
suppose
a in field R; then
reconsider z = a as Element of X by F0;
take z;
thus R reduces z,x & R reduces z,y by A2;
end;
suppose
not a in field R; then
a = x & a = y by A2,REWRITE1:18;
hence thesis;
end;
end;
theorem Ch12:
x >><< y iff x,y are_convergent_wrt the reduction of X
proof set R = the reduction of X;
thus x >><< y implies x,y are_convergent_wrt R
proof
given z such that
A1: z <=*= x & y =*=> z;
take z;
thus R reduces x,z & R reduces y,z by A1;
end;
set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
given a being object such that
A2: R reduces x,a & R reduces y,a;
per cases;
suppose
a in field R; then
reconsider z = a as Element of X by F0;
take z;
thus R reduces x,z & R reduces y,z by A2;
end;
suppose
not a in field R; then
a = x & a = y by A2,REWRITE1:18;
hence thesis;
end;
end;
theorem
x <<01>> y iff x,y are_divergent<=1_wrt the reduction of X
proof set R = the reduction of X;
thus x <<01>> y implies x,y are_divergent<=1_wrt R
proof
given z such that
A1: x <=01= z & z =01=> y;
take z;
(z ==> x or z = x) & (z ==> y or z = y) by A1;
hence ([z,x] in R or z = x) & ([z,y] in R or z = y);
end;
set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
given a being object such that
A2: ([a,x] in R or a = x) & ([a,y] in R or a = y);
a in field R or a = x or a = y by A2,RELAT_1:15; then
reconsider z = a as Element of X by F0;
take z;
thus z = x or z ==> x by A2;
thus z = y or z ==> y by A2;
end;
theorem Ch14:
x >>01<< y iff x,y are_convergent<=1_wrt the reduction of X
proof set R = the reduction of X;
thus x >>01<< y implies x,y are_convergent<=1_wrt R
proof
given z such that
A1: z <=01= x & y =01=> z;
take z;
(x ==> z or z = x) & (y ==> z or z = y) by A1;
hence ([x,z] in R or x = z) & ([y,z] in R or y = z);
end;
set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
given a being object such that
A2: ([x,a] in R or x = a) & ([y,a] in R or y = a);
a in field R or a = x or a = y by A2,RELAT_1:15; then
reconsider z = a as Element of X by F0;
take z;
thus x = z or x ==> z by A2;
thus y = z or y ==> z by A2;
end;
definition
let X;
attr X is DIAMOND means
x <<01>> y implies x >>01<< y;
attr X is CONF means
x <<>> y implies x >><< y;
attr X is CR means
x <=*=> y implies x >><< y;
attr X is WCR means
x <<01>> y implies x >><< y;
end;
definition
let X;
attr X is COMP means
X is SN CONF;
end;
scheme isCR{X() -> non empty ARS, F(Element of X()) -> Element of X()}:
X() is CR
provided
A1: for x being Element of X() holds x =*=> F(x)
and
A2: for x,y being Element of X() st x <=*=> y holds F(x) = F(y)
proof
let x,y be Element of X(); assume x <=*=> y; then
A3: F(x) = F(y) by A2;
take z = F(x);
thus thesis by A3,A1;
end;
Lm3:
x =*=> y implies x <=*=> y
proof
assume
A1: x =*=> y;
defpred P[Element of X] means x <=*=> $1;
A2: P[x];
A3: for y,z st y ==> z & P[y] holds P[z]
proof
let y,z;
assume
A4: y ==> z;
assume
A5: P[y];
A6: y <==> z by A4;
A7: y <=*=> z by A6,Th6;
thus P[z] by A5,A7,Th7;
end;
P[y] from Star1(A1,A2,A3);
hence thesis;
end;
Lm2: x <<>> y implies x <=*=> y
proof
assume
A1: x <<>> y;
consider u such that
A2: x <=*= u & u =*=> y by A1;
A3: x <=*=> u & u <=*=> y by A2,Lm3;
thus x <=*=> y by A3,Th7;
end;
Lm1: X is CR implies X is CONF by Lm2;
scheme isCOMP{X() -> non empty ARS, F(Element of X()) -> Element of X()}:
X() is COMP
provided
A1: X() is SN
and
A2: for x being Element of X() holds x =*=> F(x)
and
A3: for x,y being Element of X() st x <=*=> y holds F(x) = F(y)
proof
X() is CR from isCR(A2,A3);
hence X() is SN CONF by A1,Lm1;
end;
theorem Lem18:
x <<01>> y implies x <<>> y
proof
given z such that
A2: x <=01= z & z =01=> y;
take z;
thus x <=*= z & z =*=> y by A2,Lem1;
end;
theorem Lem18a:
x >>01<< y implies x >><< y
proof
given z such that
A2: x =01=> z & z <=01= y;
take z;
thus x =*=> z & z <=*= y by A2,Lem1;
end;
theorem
x ==> y implies x <<01>> y
proof
assume
A1: x ==> y;
take x;
thus x <=01= x & x =01=> y by A1;
end;
theorem Th17:
x ==> y implies x >>01<< y
proof
assume
A1: x ==> y;
take y;
thus x =01=> y & y =01=> y by A1;
end;
theorem
x =01=> y implies x <<01>> y;
theorem
x =01=> y implies x >>01<< y;
theorem
x <==> y implies x <<01>> y
proof
assume
A1: x <==> y;
per cases by A1;
suppose
A2: x ==> y;
take x;
thus x <=01= x & x =01=> y by A2;
end;
suppose
A3: x <== y;
take y;
thus x <=01= y & y =01=> y by A3;
end;
end;
theorem
x <==> y implies x >>01<< y
proof
assume
A1: x <==> y;
per cases by A1;
suppose
A2: x ==> y;
take y;
thus x =01=> y & y <=01= y by A2;
end;
suppose
A3: x <== y;
take x;
thus x =01=> x & x <=01= y by A3;
end;
end;
theorem
x <=01=> y implies x <<01>> y
proof
assume
A1: x <=01=> y;
per cases by A1,Lem31;
suppose x = y;
hence thesis;
end;
suppose
A2: x ==> y;
take x;
thus x <=01= x & x =01=> y by A2;
end;
suppose
A3: x <== y;
take y;
thus x <=01= y & y =01=> y by A3;
end;
end;
theorem
x <=01=> y implies x >>01<< y
proof
assume
A1: x <=01=> y;
per cases by A1,Lem31;
suppose x = y;
hence thesis;
end;
suppose
A2: x ==> y;
take y;
thus x =01=> y & y <=01= y by A2;
end;
suppose
A3: x <== y;
take x;
thus x =01=> x & x <=01= y by A3;
end;
end;
theorem Th17a:
x ==> y implies x >><< y by Th17,Lem18a;
theorem Lem17:
x =*=> y implies x >><< y;
theorem
x =*=> y implies x <<>> y;
theorem
x =+=> y implies x >><< y
proof
assume
A1: x =+=> y;
take y; thus thesis by A1,Lem2;
end;
theorem
x =+=> y implies x <<>> y
proof
assume
A1: x =+=> y;
take x; thus thesis by A1,Lem2;
end;
theorem Lm11:
x ==> y & x ==> z implies y <<01>> z
proof
assume
A1: x ==> y;
assume
A2: x ==> z;
take x;
thus y <=01= x by A1;
thus x =01=> z by A2;
end;
theorem
x ==> y & z ==> y implies x >>01<< z
proof
assume
A1: x ==> y;
assume
A2: z ==> y;
take y;
thus y <=01= x by A1;
thus z =01=> y by A2;
end;
theorem
x >><< z & z <== y implies x >><< y
proof
given u such that
A3: x =*=> u & u <=*= z;
assume
A2: z <== y;
take u;
thus x =*=> u by A3;
thus y =*=> u by A2,A3,Lem5;
end;
theorem
x >><< z & z <=01= y implies x >><< y
proof
given u such that
A3: x =*=> u & u <=*= z;
assume
A2: z <=01= y;
take u;
thus x =*=> u by A3;
thus y =*=> u by A2,A3,Lem8;
end;
theorem Lm5:
x >><< z & z <=*= y implies x >><< y
proof
given u such that
A3: x =*=> u & u <=*= z;
assume
A2: z <=*= y;
take u;
thus x =*=> u by A3;
thus y =*=> u by A2,A3,Th3;
end;
theorem Lem19:
x <<>> y implies x <=*=> y
proof
given u such that
A2: x <=*= u & u =*=> y;
A3: x <=*=> u & u <=*=> y by A2,LemZ;
thus x <=*=> y by A3,Th7;
end;
theorem
x >><< y implies x <=*=> y
proof
given u such that
A2: x =*=> u & u <=*= y;
A3: x <=*=> u & u <=*=> y by A2,LemZ;
thus x <=*=> y by A3,Th7;
end;
begin :: Church-Rosser Property
theorem
X is DIAMOND iff the reduction of X is subcommutative
proof
set R = the reduction of X;
set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
thus X is DIAMOND implies R is subcommutative
proof assume
A1: x <<01>> y implies x >>01<< y;
let a,b,c be object;
assume
A2: [a,b] in R & [a,c] in R; then
a in field R & b in field R & c in field R by RELAT_1:15; then
reconsider x = a, y = b, z = c as Element of X by F0;
x ==> y & x ==> z by A2; then
x =01=> y & x =01=> z; then
y <<01>> z;
hence b,c are_convergent<=1_wrt R by A1,Ch14;
end;
assume
A3: for a,b,c being object st [a,b] in R & [a,c] in R
holds b,c are_convergent<=1_wrt R;
let x,y; given z such that
A4: x <=01= z & z =01=> y;
per cases by A4;
suppose
x <== z & z ==> y;
hence thesis by A3,Ch14;
end;
suppose
x = z & z = y;
hence thesis;
end;
suppose
x <== z & z = y;
hence thesis by Th17;
end;
suppose
x = z & z ==> y;
hence thesis by Th17;
end;
end;
theorem Ch17:
X is CONF iff the reduction of X is confluent
proof
set R = the reduction of X;
set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
thus X is CONF implies R is confluent
proof assume
A1: x <<>> y implies x >><< y;
let a,b be object; assume
A2: a,b are_divergent_wrt R; then
A3: a,b are_convertible_wrt R by REWRITE1:37;
per cases by A3,REWRITE1:32;
suppose
a in field R & b in field R; then
reconsider x = a, y = b as Element of X by F0;
x <<>> y by A2,Ch11;
hence a,b are_convergent_wrt R by A1,Ch12;
end;
suppose
a = b;
hence a,b are_convergent_wrt R by REWRITE1:38;
end;
end;
assume
A5: for a,b being object st a,b are_divergent_wrt R
holds a,b are_convergent_wrt R;
let x,y; assume x <<>> y; then
x,y are_divergent_wrt R by Ch11;
hence thesis by A5,Ch12;
end;
theorem
X is CR iff the reduction of X is with_Church-Rosser_property
proof
set R = the reduction of X;
set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
thus X is CR implies R is with_Church-Rosser_property
proof assume
A1: x <=*=> y implies x >><< y;
let a,b be object; assume
A2: a,b are_convertible_wrt R;
per cases by A2,REWRITE1:32;
suppose
a in field R & b in field R; then
reconsider x = a, y = b as Element of X by F0;
x <=*=> y by A2;
hence a,b are_convergent_wrt R by A1,Ch12;
end;
suppose
a = b;
hence a,b are_convergent_wrt R by REWRITE1:38;
end;
end;
assume
A5: for a,b being object st a,b are_convertible_wrt R
holds a,b are_convergent_wrt R;
let x,y; assume x <=*=> y;
hence thesis by A5,Ch12;
end;
theorem
X is WCR iff the reduction of X is locally-confluent
proof
set R = the reduction of X;
set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
thus X is WCR implies R is locally-confluent
proof assume
A1: x <<01>> y implies x >><< y;
let a,b,c be object; assume
A2: [a,b] in R & [a,c] in R; then
a in field R & b in field R & c in field R by RELAT_1:15; then
reconsider x = a, y = b, z = c as Element of X by F0;
x ==> y & x ==> z by A2; then
x =01=> y & x =01=> z; then
y <<01>> z;
hence b,c are_convergent_wrt R by A1,Ch12;
end;
assume
A3: for a,b,c being object st [a,b] in R & [a,c] in R
holds b,c are_convergent_wrt R;
let x,y; given z such that
A4: x <=01= z & z =01=> y;
per cases by A4;
suppose
x <== z & z ==> y;
hence thesis by A3,Ch12;
end;
suppose
x = z & z = y;
hence thesis;
end;
suppose
x <== z & z = y;
hence thesis by Th17a;
end;
suppose
x = z & z ==> y;
hence thesis by Th17a;
end;
end;
theorem
for X being non empty ARS holds X is COMP iff the reduction of X is complete
proof let X be non empty ARS;
set R = the reduction of X;
A2: X is CONF iff R is confluent by Ch17;
X is SN iff R is strongly-normalizing by Ch7,Ch8;
hence thesis by A2;
end;
theorem LemA:
X is DIAMOND & x <=*= z & z =01=> y implies
ex u st x =01=> u & u <=*= y
proof
assume
A1: for x,y st x <<01>> y holds x >>01<< y;
assume
A2: x <=*= z;
assume
A3: z =01=> y;
defpred P[Element of X] means ex u st $1 =01=> u & u <=*= y;
A4: for u,v st u ==> v & P[u] holds P[v]
proof
let u,v;
assume u ==> v; then
B1: u =01=> v;
given w such that
B2: u =01=> w & w <=*= y;
v <<01>> w by B1,B2; then
v >>01<< w by A1; then
consider u such that
B3: v =01=> u & u <=01= w;
thus P[v] by B2,B3,Lem11;
end;
A5: for u,v st u =*=> v & P[u] holds P[v] from Star(A4);
thus thesis by A5,A2,A3;
end;
theorem
X is DIAMOND & x <=01= y & y =*=> z implies
ex u st x =*=> u & u <=01= z
proof
assume X is DIAMOND & x <=01= y & y =*=> z; then
ex u st z =01=> u & u <=*= x by LemA;
hence thesis;
end;
registration
cluster DIAMOND -> CONF for ARS;
coherence
proof let X;
assume
A1: X is DIAMOND;
let x,y;
given z such that
A2: x <=*= z and
A3: z =*=> y;
defpred P[Element of X] means x >><< $1;
A4: P[z] by A2,Lem17;
A5: for u,v st u ==> v & P[u] holds P[v]
proof
let u,v;
assume
A6: u ==> v;
given w such that
A7: x =*=> w & w <=*= u;
A8: u =01=> v by A6;
consider a such that
A9: w =01=> a & a <=*= v by A1,A7,A8,LemA;
A10: x =*=> a by A7,A9,Lem11;
thus P[v] by A9,A10,DEF2;
end;
P[y] from Star1(A3,A4,A5);
hence x >><< y;
end;
end;
registration
cluster DIAMOND -> CR for ARS;
coherence
proof let X;
assume
A1: X is DIAMOND;
let x,y;
assume
A2: x <=*=> y;
defpred P[Element of X] means x >><< $1;
A4: P[x];
A5: for u,v st u <==> v & P[u] holds P[v]
proof
let u,v;
assume
A6: u <==> v;
given w such that
A7: x =*=> w & w <=*= u;
per cases by A6;
suppose u ==> v; then
A8: u =01=> v;
consider a such that
A9: w =01=> a & a <=*= v by A1,A7,A8,LemA;
A10: x =*=> a by A7,A9,Lem11;
thus P[v] by A9,A10,DEF2;
end;
suppose u <== v; then
A11: v =*=> w by A7,Lem5;
thus P[v] by A7,A11,DEF2;
end;
end;
P[y] from Star2A(A2,A4,A5);
hence x >><< y;
end;
end;
registration
cluster CR -> WCR for ARS;
coherence
proof let X;
assume
A1: X is CR;
let x,y;
assume
A2: x <<01>> y;
A4: x <=*=> y by A2,Lem18,Lem19;
thus x >><< y by A1,A4;
end;
end;
registration
cluster CR -> CONF for ARS;
coherence by Lm1;
end;
registration
cluster CONF -> CR for ARS;
coherence
proof let X;
assume
A1: X is CONF;
let x;
defpred P[Element of X] means x >><< $1;
A3: for y,z st y <==> z & P[y] holds P[z]
proof
let y,z;
assume
B1: y <==> z & P[y];
consider u such that
B2: x =*=> u & u <=*= y by B1,DEF2;
per cases by B1;
suppose
B3: y ==> z;
y =*=> z by B3,Th2; then
u <<>> z by B2;
hence P[z] by A1,B2,Lm5;
end;
suppose
B5: y <== z;
thus P[z] by B1,B5,Th2,Lm5;
end;
end;
for y,z st y <=*=> z & P[y] holds P[z] from Star2(A3);
hence thesis;
end;
end;
theorem
X is non CONF WN implies
ex x,y,z st y is_normform_of x & z is_normform_of x & y <> z
proof
given a,b such that
A1: a <<>> b & not a >><< b;
consider x such that
A0: a <=*= x & x =*=> b by A1;
assume
A2: c is normalizable; then
a is normalizable; then
consider y such that
A3: y is_normform_of a;
b is normalizable by A2; then
consider z such that
A4: z is_normform_of b;
take x,y,z;
thus y is_normform_of x & z is_normform_of x by A0,A3,A4,LemN7;
thus thesis by A1,A3,A4;
end;
registration
::$N Newman's lemma
cluster SN WCR -> CR for ARS;
coherence
proof let X;
assume
A1: X is SN WCR;
assume
A2: X is not CR;
A3: X is not CONF by A2;
consider x1,x2 being Element of X such that
A4: x1 <<>> x2 & not x1 >><< x2 by A3;
defpred P[Element of X] means
ex x,y st x is_normform_of $1 & y is_normform_of $1 & x <> y;
A5: ex x st P[x]
proof
consider x such that
B1: x1 <=*= x & x =*=> x2 by A4;
take x;
consider y1 being Element of X such that
B2: y1 is_normform_of x1 by A1,ThWN1;
consider y2 being Element of X such that
B3: y2 is_normform_of x2 by A1,ThWN1;
take y1,y2;
thus y1 is_normform_of x by B1,B2,LemN7;
thus y2 is_normform_of x by B1,B3,LemN7;
assume
B4: y1 = y2;
thus contradiction by A4,B2,B3,B4;
end;
A6: for x st P[x] ex y st P[y] & x ==> y
proof
let x;
assume P[x]; then
consider x1,x2 being Element of X such that
C1: x1 is_normform_of x & x2 is_normform_of x & x1 <> x2;
x =+=> x1 by C1,Lem21; then
consider y1 being Element of X such that
C2: x ==> y1 & y1 =*=> x1;
x =+=> x2 by C1,Lem21; then
consider y2 being Element of X such that
C3: x ==> y2 & y2 =*=> x2;
y1 >><< y2 by A1,C2,C3,Lm11; then
consider y such that
C4: y1 =*=> y & y <=*= y2;
consider y0 being Element of X such that
C5: y0 is_normform_of y by A1,ThWN1;
per cases;
suppose
D1: y0 = x1;
take y2;
D2: y0 is_normform_of y2 by C4,C5,LemN7;
x2 is_normform_of y2 by C1,C3;
hence P[y2] by C1,D1,D2;
thus x ==> y2 by C3;
end;
suppose
D3: y0 <> x1;
take y1;
D4: y0 is_normform_of y1 by C4,C5,LemN7;
x1 is_normform_of y1 by C1,C2;
hence thesis by C2,D3,D4;
end;
end;
A7: X is not SN from notSN(A5,A6);
thus contradiction by A1,A7;
end;
end;
registration
cluster CR -> N.F. for ARS;
coherence
proof let X;
assume
A1: X is CR;
let x,y;
assume
A2: x is normform;
assume
A3: x <=*=> y;
A4: x >><< y by A1,A3;
consider z such that
A5: x =*=> z & z <=*= y by A4;
thus y =*=> x by A2,A5,LemN1;
end;
end;
registration
cluster WN UN -> CR for ARS;
coherence
proof let X;
assume
A1: X is WN;
assume
A2: X is UN;
let x,y;
assume
A3: x <=*=> y;
A4: x is normalizable & y is normalizable by A1;
consider u such that
A5: u is_normform_of x by A4;
consider v such that
A6: v is_normform_of y by A4;
A7: u is normform & x =*=> u by A5;
take u;
thus x =*=> u by A5;
u <=*=> x by A5,LemZ; then
u <=*=> y & y <=*=> v by A3,A6,Th7,LemZ;
hence y =*=> u by A2,A7,A6,Th7;
end;
end;
registration
cluster SN CR -> COMP for ARS;
coherence;
cluster COMP -> CR WCR N.F. UN UN* WN for ARS;
coherence;
end;
theorem
X is COMP implies
for x,y st x <=*=> y holds nf x = nf y by Lem27;
registration
cluster WN UN* -> CR for ARS;
coherence;
cluster SN UN* -> COMP for ARS;
coherence;
end;
begin :: Term Rewriting Systems
definition
struct(ARS,UAStr) TRSStr (#
carrier -> set,
charact -> PFuncFinSequence of the carrier,
reduction -> Relation of the carrier
#);
end;
registration
cluster non empty non-empty strict for TRSStr;
existence
proof
set S = the non empty set;
set o = the non-empty non empty PFuncFinSequence of S;
set r = the Relation of S;
take X = TRSStr(#S, o, r#);
thus the carrier of X is non empty;
thus the charact of X <> {};
thus thesis;
end;
end;
definition
let S be non empty UAStr;
attr S is Group-like means
Seg 3 c= dom the charact of S &
for f being non empty homogeneous
PartFunc of (the carrier of S)*, the carrier of S holds
(f = (the charact of S).1 implies arity f = 0) &
(f = (the charact of S).2 implies arity f = 1) &
(f = (the charact of S).3 implies arity f = 2);
end;
theorem Th01:
for X being non empty set
for f1,f2,f3 being non empty homogeneous PartFunc of X*, X
st arity f1 = 0 & arity f2 = 1 & arity f3 = 2
for S being non empty UAStr
st the carrier of S = X & <*f1,f2,f3*> c= the charact of S
holds S is Group-like
proof
let X be non empty set;
let f1,f2,f3 be non empty homogeneous PartFunc of X*, X;
assume
01: arity f1 = 0;
assume
02: arity f2 = 1;
assume
03: arity f3 = 2;
let S be non empty UAStr;
assume
04: the carrier of S = X & <*f1,f2,f3*> c= the charact of S;
05: dom <*f1,f2,f3*> = Seg 3 by FINSEQ_2:124;
hence Seg 3 c= dom the charact of S by 04,RELAT_1:11;
let f be non empty homogeneous
PartFunc of (the carrier of S)*, the carrier of S;
1 in Seg 3 & 2 in Seg 3 & 3 in Seg 3 by FINSEQ_3:1,ENUMSET1:def 1; then
(the charact of S).1 = <*f1,f2,f3*>.1 &
(the charact of S).2 = <*f1,f2,f3*>.2 &
(the charact of S).3 = <*f1,f2,f3*>.3 by 04,05,GRFUNC_1:2;
hence (f = (the charact of S).1 implies arity f = 0) &
(f = (the charact of S).2 implies arity f = 1) &
(f = (the charact of S).3 implies arity f = 2) by 01,02,03,FINSEQ_1:45;
end;
theorem Th02:
for X being non empty set
for f1,f2,f3 being non empty quasi_total homogeneous PartFunc of X*, X
for S being non empty UAStr
st the carrier of S = X & <*f1,f2,f3*> = the charact of S
holds S is quasi_total partial
proof
let X be non empty set;
let f1,f2,f3 be non empty quasi_total homogeneous PartFunc of X*, X;
let S be non empty UAStr;
assume
04: the carrier of S = X & <*f1,f2,f3*> = the charact of S;
set A = the carrier of S;
thus S is quasi_total
proof
let i be Nat, h being PartFunc of A*,A;
assume i in dom the charact of S; then
i in Seg 3 by 04,FINSEQ_1:89; then
i = 1 or i = 2 or i = 3 by FINSEQ_3:1,ENUMSET1:def 1;
hence thesis by 04,FINSEQ_1:45;
end;
let i be Nat, h being PartFunc of A*,A;
assume i in dom the charact of S; then
i in Seg 3 by 04,FINSEQ_1:89; then
i = 1 or i = 2 or i = 3 by FINSEQ_3:1,ENUMSET1:def 1;
hence thesis by 04,FINSEQ_1:45;
end;
definition
let S be non empty non-empty UAStr;
let o be operation of S;
let a be Element of dom o;
redefine func o.a -> Element of S;
coherence
proof
o in rng the charact of S; then
o <> {} & o in PFuncs((the carrier of S)*, the carrier of S)
by RELAT_1:def 9; then
o.a in rng o & rng o c= the carrier of S by RELAT_1:def 19,FUNCT_1:3;
hence thesis;
end;
end;
registration
let S be non empty non-empty UAStr;
cluster -> non empty for operation of S;
coherence by RELAT_1:def 9;
let o be operation of S;
cluster -> Relation-like Function-like for Element of dom o;
coherence
proof
let a be Element of dom o;
a in dom o & dom o c= (the carrier of S)*; then
a is Element of (the carrier of S)*;
hence thesis;
end;
end;
registration
let S be partial non empty non-empty UAStr;
cluster -> homogeneous for operation of S;
coherence
proof
let o be operation of S;
consider i being object such that
A1: i in dom the charact of S & o = (the charact of S).i by FUNCT_1:def 3;
thus thesis by A1;
end;
end;
registration
let S be quasi_total non empty non-empty UAStr;
cluster -> quasi_total for operation of S;
coherence
proof
let o be operation of S;
consider i being object such that
A1: i in dom the charact of S & o = (the charact of S).i by FUNCT_1:def 3;
thus thesis by A1,MARGREL1:def 24;
end;
end;
theorem ThA:
for S being non empty non-empty UAStr st S is Group-like
holds
1 is OperSymbol of S & 2 is OperSymbol of S & 3 is OperSymbol of S
proof
let S be non empty non-empty UAStr;
assume
A0: Seg 3 c= dom the charact of S;
1 in Seg 3 & 2 in Seg 3 & 3 in Seg 3 by FINSEQ_3:1,ENUMSET1:def 1;
hence thesis by A0;
end;
theorem ThB:
for S being partial non empty non-empty UAStr st S is Group-like
holds
arity Den(In(1, dom the charact of S), S) = 0 &
arity Den(In(2, dom the charact of S), S) = 1 &
arity Den(In(3, dom the charact of S), S) = 2
proof
let S be partial non empty non-empty UAStr;
assume
A1: S is Group-like; then
1 is OperSymbol of S & 2 is OperSymbol of S & 3 is OperSymbol of S
by ThA; then
In(1, dom the charact of S) = 1 &
In(2, dom the charact of S) = 2 &
In(3, dom the charact of S) = 3;
hence thesis by A1,PUA2MSS1:def 1;
end;
definition
let S be non empty non-empty TRSStr;
attr S is invariant means:
DEF2:
for o being OperSymbol of S
for a,b being Element of dom Den(o,S)
for i being Nat st i in dom a
for x,y being Element of S
st x = a.i & b = a+*(i,y) & x ==> y
holds Den(o,S).a ==> Den(o,S).b;
end;
definition
let S be non empty non-empty TRSStr;
attr S is compatible means
for o being OperSymbol of S
for a,b being Element of dom Den(o,S)
st for i being Nat st i in dom a holds
for x,y being Element of S st x = a.i & y = b.i holds x ==> y
holds Den(o,S).a =*=> Den(o,S).b;
end;
theorem Th0:
for n being natural number, X being non empty set, x being Element of X
ex f being non empty homogeneous quasi_total PartFunc of X*, X st
arity f = n & f = (n-tuples_on X) --> x
proof
let n be natural number, X be non empty set;
let x be Element of X;
set f = (n-tuples_on X) --> x;
A1: dom f = n-tuples_on X & rng f = {x} & n in omega
by FUNCOP_1:8,ORDINAL1:def 12; then
dom f c= X* & rng f c= X by ZFMISC_1:31,FINSEQ_2:134; then
reconsider f as non empty PartFunc of X*, X by RELSET_1:4;
A2: f is quasi_total
proof
let x,y be FinSequence of X; assume
len x = len y & x in dom f; then
len x = n & len y = n by A1,FINSEQ_2:132;
hence thesis by FINSEQ_2:133;
end;
f is homogeneous
proof
let x,y be FinSequence; assume
x in dom f & y in dom f; then
reconsider x,y as Element of n-tuples_on X;
len x = n & len y = n by A1,FINSEQ_2:132;
hence thesis;
end; then
reconsider f as non empty homogeneous quasi_total PartFunc of X*, X by A2;
take f;
set y = the Element of n-tuples_on X;
A3: for x being FinSequence st x in dom f holds n = len x by A1,FINSEQ_2:132;
y in dom f;
hence arity f = n by A3,MARGREL1:def 25;
thus thesis;
end;
registration
let X be non empty set;
let O be PFuncFinSequence of X;
let r be Relation of X;
cluster TRSStr(#X, O, r#) -> non empty;
coherence;
end;
registration
let X be non empty set;
let O be non empty non-empty PFuncFinSequence of X;
let r be Relation of X;
cluster TRSStr(#X, O, r#) -> non-empty;
coherence
proof
thus the charact of TRSStr(#X, O, r#) <> {};
thus the charact of TRSStr(#X, O, r#) is non-empty;
end;
end;
definition
let X be non empty set;
let x be Element of X;
func TotalTRS(X,x) -> non empty non-empty strict TRSStr means:
DEF3:
the carrier of it = X &
the charact of it =
<*(0-tuples_on X)-->x, (1-tuples_on X)-->x, (2-tuples_on X)-->x*> &
the reduction of it = nabla X;
uniqueness;
existence
proof
consider f0 being non empty homogeneous quasi_total PartFunc of X*, X
such that
A0: arity f0 = 0 & f0 = (0-tuples_on X) --> x by Th0;
consider f1 being non empty homogeneous quasi_total PartFunc of X*, X
such that
A1: arity f1 = 1 & f1 = (1-tuples_on X) --> x by Th0;
consider f2 being non empty homogeneous quasi_total PartFunc of X*, X
such that
A2: arity f2 = 2 & f2 = (2-tuples_on X) --> x by Th0;
set r = nabla X;
reconsider a = f0, b = f1, c = f2 as Element of PFuncs(X*, X)
by PARTFUN1:45;
reconsider O = <*a,b,c*> as non empty non-empty PFuncFinSequence of X;
take S = TRSStr(#X, O, r#);
thus thesis by A0,A1,A2;
end;
end;
registration
let X be non empty set;
let x be Element of X;
cluster TotalTRS(X,x) -> quasi_total partial Group-like invariant;
coherence
proof set S = TotalTRS(X,x);
A3: the carrier of S = X & the charact of S =
<*(0-tuples_on X)-->x, (1-tuples_on X)-->x, (2-tuples_on X)-->x*> &
the reduction of S = nabla X by DEF3;
consider f0 being non empty homogeneous quasi_total PartFunc of X*, X
such that
A0: arity f0 = 0 & f0 = (0-tuples_on X) --> x by Th0;
consider f1 being non empty homogeneous quasi_total PartFunc of X*, X
such that
A1: arity f1 = 1 & f1 = (1-tuples_on X) --> x by Th0;
consider f2 being non empty homogeneous quasi_total PartFunc of X*, X
such that
A2: arity f2 = 2 & f2 = (2-tuples_on X) --> x by Th0;
[:X,X:] c= [:X,X:]; then
reconsider r = [:X,X:] as Relation of X;
reconsider a = f0, b = f1, c = f2 as Element of PFuncs(X*, X)
by PARTFUN1:45;
thus S is quasi_total partial Group-like by A0,A1,A2,A3,Th01,Th02;
let o be OperSymbol of S;
let a,b be Element of dom Den(o,S);
let i be Nat such that i in dom a;
let x,y be Element of S such that x = a.i & b = a+*(i,y) & x ==> y;
thus [Den(o,S).a,Den(o,S).b] in the reduction of S by A3,ZFMISC_1:87;
end;
end;
registration
cluster strict quasi_total partial Group-like invariant for
non empty non-empty TRSStr;
existence
proof
take TotalTRS(NAT,In(0,NAT));
thus thesis;
end;
end;
definition
let S be Group-like quasi_total partial non empty non-empty TRSStr;
func 1.S -> Element of S equals
Den(In(1,dom the charact of S), S).{};
coherence
proof
arity Den(In(1,dom the charact of S), S) = 0 by ThB; then
dom Den(In(1,dom the charact of S), S) = 0-tuples_on the carrier of S
by COMPUT_1:22 .= {{}} by COMPUT_1:5; then
{} in dom Den(In(1,dom the charact of S), S) by TARSKI:def 1;
hence thesis by FUNCT_1:102;
end;
let a be Element of S;
func a " -> Element of S equals
Den(In(2,dom the charact of S), S).<*a*>;
coherence
proof
arity Den(In(2,dom the charact of S), S) = 1 by ThB; then
dom Den(In(2,dom the charact of S), S) = 1-tuples_on the carrier of S &
<*a*> is Element of 1-tuples_on the carrier of S
by FINSEQ_2:98,MARGREL1:22;
hence thesis by FUNCT_1:102;
end;
let b be Element of S;
func a * b -> Element of S equals
Den(In(3,dom the charact of S), S).<*a,b*>;
coherence
proof
arity Den(In(3,dom the charact of S), S) = 2 by ThB; then
dom Den(In(3,dom the charact of S), S) = 2-tuples_on the carrier of S &
<*a,b*> is Element of 2-tuples_on the carrier of S
by FINSEQ_2:101,MARGREL1:22;
hence thesis by FUNCT_1:102;
end;
end;
reserve
S for Group-like quasi_total partial invariant non empty non-empty TRSStr;
reserve a,b,c for Element of S;
theorem
a ==> b implies a" ==> b"
proof
assume
A0: a ==> b;
set o = In(2, dom the charact of S);
arity Den(o, S) = 1 by ThB; then
dom Den(o, S) = 1-tuples_on the carrier of S by MARGREL1:22; then
reconsider aa = <*a*>, bb = <*b*> as Element of dom Den(o, S)
by FINSEQ_2:98;
A2: dom <*a*> = Seg 1 & 1 in Seg 1 by FINSEQ_1:1,38;
A3: <*a*>.1 = a by FINSEQ_1:40;
<*a*>+*(1,b) = <*b*> by FUNCT_7:95; then
Den(o,S).aa ==> Den(o,S).bb by A0,A2,A3,DEF2;
hence a" ==> b";
end;
theorem ThI2:
a ==> b implies a*c ==> b*c
proof
assume
A0: a ==> b;
set o = In(3, dom the charact of S);
arity Den(o, S) = 2 by ThB; then
dom Den(o, S) = 2-tuples_on the carrier of S by MARGREL1:22; then
reconsider ac = <*a,c*>, bc = <*b,c*> as Element of dom Den(o, S)
by FINSEQ_2:101;
A2: dom <*a,c*> = Seg 2 & 1 in Seg 2 by FINSEQ_1:1,89;
A3: <*a,c*>.1 = a by FINSEQ_1:44;
<*a,c*>+*(1,b) = <*b,c*> by COMPUT_1:1; then
Den(o,S).ac ==> Den(o,S).bc by A0,A2,A3,DEF2;
hence a*c ==> b*c;
end;
theorem ThI3:
a ==> b implies c*a ==> c*b
proof
assume
A0: a ==> b;
set o = In(3, dom the charact of S);
arity Den(o, S) = 2 by ThB; then
dom Den(o, S) = 2-tuples_on the carrier of S by MARGREL1:22; then
reconsider ac = <*c,a*>, bc = <*c,b*> as Element of dom Den(o, S)
by FINSEQ_2:101;
A2: dom <*c,a*> = Seg 2 & 2 in Seg 2 by FINSEQ_1:1,89;
A3: <*c,a*>.2 = a by FINSEQ_1:44;
<*c,a*>+*(2,b) = <*c,b*> by COMPUT_1:1; then
Den(o,S).ac ==> Den(o,S).bc by A0,A2,A3,DEF2;
hence c*a ==> c*b;
end;
begin :: An Execution of Knuth-Bendix Algorithm
reserve S for Group-like quasi_total partial non empty non-empty TRSStr;
reserve a,b,c for Element of S;
definition
let S;
attr S is (R1) means
1.S * a ==> a;
attr S is (R2) means
a" * a ==> 1.S;
attr S is (R3) means
(a * b) * c ==> a * (b * c);
attr S is (R4) means
a" * (a * b) ==> b;
attr S is (R5) means
(1.S)" * a ==> a;
attr S is (R6) means
(a")" * 1.S ==> a;
attr S is (R7) means
(a")" * b ==> a * b;
attr S is (R8) means
a * 1.S ==> a;
attr S is (R9) means
(a")" ==> a;
attr S is (R10) means
(1.S)" ==> 1.S;
attr S is (R11) means
a * (a") ==> 1.S;
attr S is (R12) means
a * (a" * b) ==> b;
attr S is (R13) means
a * (b * (a * b)") ==> 1.S;
attr S is (R14) means
a * (b * a)" ==> b";
attr S is (R15) means
(a * b)" ==> b" * a";
end;
reserve
S for Group-like quasi_total partial invariant non empty non-empty TRSStr,
a,b,c for Element of S;
theorem
S is (R1) (R2) (R3) implies a" * (a * b) <<>> b
proof
assume
A1: S is (R1) (R2) (R3);
take (a"*a)*b;
thus (a"*a)*b =*=> a"*(a*b) by A1,Th2;
(a"*a)*b ==> 1.S * b & 1.S * b ==> b by A1,ThI2; then
(a"*a)*b =*=> 1.S * b & 1.S * b =*=> b by Th2;
hence (a"*a)*b =*=> b by Th3;
end;
theorem
S is (R1) (R4) implies (1.S)" * a <<>> a
proof
assume
A1: S is (R1) (R4);
take (1.S)"*(1.S*a);
1.S*a ==> a by A1;
hence (1.S)"*(1.S*a) =*=> (1.S)" * a by Th2,ThI3;
thus thesis by A1,Th2;
end;
theorem
S is (R2) (R4) implies (a")" * 1.S <<>> a
proof
assume
A1: S is (R2) (R4);
take (a")" * (a" * a);
a" * a ==> 1.S by A1;
hence (a")" * (a" * a) =*=> (a")" * 1.S by Th2,ThI3;
thus (a")" * (a" * a) =*=> a by A1,Th2;
end;
theorem
S is (R1) (R3) (R6) implies (a")" * b <<>> a * b
proof
assume
A1: S is (R1) (R3) (R6);
take (a""*1.S)*b;
A2: (a""*1.S)*b =*=> a""*(1.S*b) by A1,Th2;
1.S*b ==> b by A1; then
a""*(1.S*b) =*=> a""*b by Th2,ThI3;
hence (a""*1.S)*b =*=> a""*b by A2,Th3;
a"" * 1.S ==> a by A1;
hence (a"" * 1.S) * b =*=> a * b by Th2,ThI2;
end;
theorem
S is (R6) (R7) implies a * 1.S <<>> a
proof
assume
A1: S is (R6) (R7);
take a""*1.S;
thus a""*1.S =*=> a*1.S by A1,Th2;
thus a"" * 1.S =*=> a by A1,Th2;
end;
theorem
S is (R6) (R8) implies (a")" <<>> a
proof
assume
A1: S is (R6) (R8);
take a""*1.S;
thus a""*1.S =*=> a"" by A1,Th2;
thus a"" * 1.S =*=> a by A1,Th2;
end;
theorem
S is (R5) (R8) implies (1.S)" <<>> 1.S
proof
assume
A1: S is (R5) (R8);
take (1.S)"*1.S;
thus (1.S)"*1.S =*=> (1.S)" by A1,Th2;
thus (1.S)" * 1.S =*=> 1.S by A1,Th2;
end;
theorem
S is (R2) (R9) implies a * (a") <<>> 1.S
proof
assume
A1: S is (R2) (R9);
take a""*a";
a"" ==> a by A1;
hence a""*a" =*=> a*a" by Th2,ThI2;
thus a""*a" =*=> 1.S by A1,Th2;
end;
theorem
S is (R1) (R3) (R11) implies a * (a" * b) <<>> b
proof
assume
A1: S is (R1) (R3) (R11);
take (a * a") * b;
thus (a * a") * b =*=> a * (a" * b) by A1,Th2;
(a * a") * b ==> 1.S * b & 1.S * b ==> b by A1,ThI2;
hence (a * a") * b =*=> b by Lem3;
end;
theorem
S is (R3) (R11) implies a * (b * (a * b)") <<>> 1.S
proof
assume
A1: S is (R3) (R11);
take (a * b) * (a * b)";
thus (a * b) * (a * b)" =*=> a * (b * (a * b)") by A1,Th2;
thus (a * b) * (a * b)" =*=> 1.S by A1,Th2;
end;
theorem
S is (R4) (R8) (R13) implies a * (b * a)" <<>> b"
proof
assume
A1: S is (R4) (R8) (R13);
take b"*(b*(a*(b*a)"));
thus b"*(b*(a*(b*a)")) =*=> a*(b*a)" by A1,Th2;
b"*(b*(a*(b*a)")) ==> b"*1.S & b"*1.S ==> b" by A1,ThI3;
hence b"*(b*(a*(b*a)")) =*=> b" by Lem3;
end;
theorem
S is (R4) (R14) implies (a * b)" <<>> b" * a"
proof
assume
A1: S is (R4) (R14);
take b"*(b*(a*b)");
thus b"*(b*(a*b)") =*=> (a * b)" by A1,Th2;
(b*(a*b)") ==> a" by A1;
hence b"*(b*(a*b)") =*=> b" * a" by Th2,ThI3;
end;
theorem
S is (R1) (R10) implies (1.S)" * a =*=> a
proof
assume
A1: S is (R1) (R10);
(1.S)"*a ==> 1.S*a & 1.S*a ==> a by A1,ThI2;
hence (1.S)" * a =*=> a by Lem3;
end;
theorem
S is (R8) (R9) implies (a")" * 1.S =*=> a
proof
assume S is (R8) (R9); then
(a")" * 1.S ==> a"" & a"" ==> a;
hence (a")" * 1.S =*=> a by Lem3;
end;
theorem
S is (R9) implies (a")" * b =*=> a * b
proof
assume S is (R9); then
a"" ==> a;
hence (a")" * b =*=> a * b by Th2,ThI2;
end;
theorem
S is (R11) (R14) implies a * (b * (a * b)") =*=> 1.S
proof
assume
A1: S is (R11) (R14);
a * (b * (a * b)") ==> a*a" & a*a" ==> 1.S by A1,ThI3;
hence a * (b * (a * b)") =*=> 1.S by Lem3;
end;
theorem
S is (R12) (R15) implies a * (b * a)" =*=> b"
proof
assume
A1: S is (R12) (R15);
a * (b * a)" ==> a*(a"*b") & a*(a"*b") ==> b" by A1,ThI3;
hence a * (b * a)" =*=> b" by Lem3;
end;