:: General Theory of Quasi-Commutative BCI-algebras
:: by Tao Sun , Weibo Pan , Chenglong Wu and Xiquan Liang
::
:: Received May 13, 2008
:: Copyright (c) 2008-2021 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 NUMBERS, BCIALG_1, SUBSET_1, POLYEQ_1, XBOOLE_0, POWER, ARYTM_3,
FUNCT_1, STRUCT_0, XXREAL_0, SUPINF_2, CARD_1, ARYTM_1, NAT_1, BINOP_1,
TARSKI, CHORD, FILTER_0, BCIALG_3, BCIALG_5;
notations TARSKI, XBOOLE_0, SUBSET_1, STRUCT_0, BCIALG_1, ORDINAL1, NUMBERS,
XXREAL_0, REAL_1, FUNCT_2, XCMPLX_0, NAT_1, BCIALG_2, BCIALG_3;
constructors REAL_1, NAT_1, BCIALG_2, BCIALG_3, XREAL_0, NUMBERS;
registrations BCIALG_1, STRUCT_0, ORDINAL1, XXREAL_0, XREAL_0, BCIALG_3,
NAT_1, BCIALG_2;
requirements SUBSET, NUMERALS, REAL, ARITHM;
definitions TARSKI, XBOOLE_0;
equalities BCIALG_1;
expansions TARSKI, BCIALG_1;
theorems TARSKI, STRUCT_0, BCIALG_1, XREAL_1, XXREAL_0, XBOOLE_0, BCIALG_3,
NAT_1, BCIALG_2;
schemes NAT_1;
begin :: The Basics of General Theory of Quasi-Commutative BCI-algebra
definition
let X be BCI-algebra;
let x,y be Element of X;
let m,n be Nat;
func Polynom (m,n,x,y) -> Element of X equals
(((x,(x\y)) to_power (m+1)),(y
\x)) to_power n;
coherence;
end;
reserve X for BCI-algebra;
reserve x,y,z for Element of X;
reserve i,j,k,l,m,n for Nat;
reserve f,g for sequence of the carrier of X;
theorem Th1:
x <= y & y <= z implies x <= z
by BCIALG_1:3;
theorem Th2:
x <= y & y <= x implies x = y
by BCIALG_1:def 7;
theorem Th3:
for X being BCK-algebra,x,y being Element of X holds x\y <= x & (
x,y) to_power (n+1) <= (x,y) to_power n
proof
let X be BCK-algebra;
let x,y be Element of X;
((x,y) to_power n\y)\(x,y) to_power n = ((x,y) to_power n\(x,y) to_power
n)\y by BCIALG_1:7
.= y` by BCIALG_1:def 5
.= 0.X by BCIALG_1:def 8;
then
A1: (x,y) to_power n\y <= (x,y) to_power n;
(x\y)\x = (x\x)\y by BCIALG_1:7
.= y` by BCIALG_1:def 5
.= 0.X by BCIALG_1:def 8;
hence thesis by A1,BCIALG_2:4;
end;
theorem Th4:
for X being BCK-algebra,x being Element of X holds (0.X,x) to_power n = 0.X
proof
let X be BCK-algebra;
let x be Element of X;
defpred P[Nat] means $1<= n implies (0.X,x) to_power $1 = 0.X;
now
let k;
assume
A1: k<= n implies (0.X,x) to_power k = 0.X;
set m = k+1;
assume m<=n;
then (0.X,x) to_power m = x` by A1,BCIALG_2:4,NAT_1:13
.= 0.X by BCIALG_1:def 8;
hence (0.X,x) to_power m = 0.X;
end;
then
A2: for k st P[k] holds P[k+1];
A3: P[0] by BCIALG_2:1;
for n holds P[n] from NAT_1:sch 2(A3,A2);
hence thesis;
end;
theorem Th5:
for X being BCK-algebra,x,y being Element of X st m>=n
holds (x,y) to_power m <= (x,y) to_power n
proof
let X be BCK-algebra;
let x,y be Element of X;
assume m>=n;
then m - n is Element of NAT by NAT_1:21;
then consider k being Element of NAT such that
A1: k=m-n;
((x,y) to_power k)\x = (x\x,y) to_power k by BCIALG_2:7
.= (0.X,y) to_power k by BCIALG_1:def 5
.= 0.X by Th4;
then (x,y) to_power k <= x;
then ((x,y) to_power k,y) to_power n <= (x,y) to_power n by BCIALG_2:19;
then (x,y) to_power (k+n) <= (x,y) to_power n by BCIALG_2:10;
hence thesis by A1;
end;
theorem Th6:
for X being BCK-algebra, x,y being Element of X st m>n & (x,y)
to_power n = (x,y) to_power m holds for k being Nat st k >=n holds (
x,y) to_power n = (x,y) to_power k
proof
let X be BCK-algebra;
let x,y be Element of X;
assume that
A1: m>n and
A2: (x,y) to_power n = (x,y) to_power m;
m - n is Element of NAT & m-n>n-n by A1,NAT_1:21,XREAL_1:9;
then m-n >=1 by NAT_1:14;
then m-n+n >= 1+n by XREAL_1:6;
then
A3: (x,y) to_power n <= (x,y) to_power (n+1) by A2,Th5;
A4: (x,y) to_power (n+1) <= (x,y) to_power n by Th3;
for k being Nat st k >=n holds (x,y) to_power n = (x,y) to_power k
proof
let k be Nat;
assume k >=n;
then k - n is Element of NAT by NAT_1:21;
then consider k1 being Element of NAT such that
A5: k1=k-n;
(x,y) to_power n = ((x,y) to_power n,y) to_power k1
proof
defpred P[Nat] means
$1<= k1 implies (x,y) to_power n = ((x,y
) to_power n,y) to_power $1;
now
let k;
assume
A6: k<= k1 implies (x,y) to_power n = ((x,y) to_power n,y) to_power k;
set m=k+1;
A7: ((x,y) to_power n,y) to_power m = ((x,y) to_power n,y) to_power k
\y by BCIALG_2:4
.= ((x,y) to_power n\y,y) to_power k by BCIALG_2:7
.= ((x,y) to_power (n+1),y) to_power k by BCIALG_2:4
.= ((x,y) to_power n,y) to_power k by A4,A3,Th2;
assume m<=k1;
hence (x,y) to_power n = ((x,y) to_power n,y) to_power m by A6,A7,
NAT_1:13;
end;
then
A8: for k st P[k] holds P[k+1];
A9: P[0] by BCIALG_2:1;
for n holds P[n] from NAT_1:sch 2(A9,A8);
hence thesis;
end;
then (x,y) to_power n = (x,y) to_power (n+k1) by BCIALG_2:10;
hence thesis by A5;
end;
hence thesis;
end;
theorem Th7:
Polynom (0,0,x,y) = x\(x\y)
proof
Polynom (0,0,x,y) = (x,(x\y)) to_power (0+1) by BCIALG_2:1
.= x\(x\y) by BCIALG_2:2;
hence thesis;
end;
theorem
Polynom (m,n,x,y) = (((Polynom (0,0,x,y),(x\y)) to_power m),(y\x)) to_power n
proof
(Polynom (0,0,x,y),(x\y)) to_power m = ((x\(x\y)),(x\y)) to_power m by Th7
.= ((x,(x\y)) to_power m) \ (x\y) by BCIALG_2:7
.= (x,(x\y)) to_power (m+1) by BCIALG_2:4;
hence thesis;
end;
theorem Th9:
Polynom (m+1,n,x,y) = Polynom (m,n,x,y) \ (x\y)
proof
Polynom (m+1,n,x,y) = (((x,(y\x)) to_power n),(x\y)) to_power (m+1+1) by
BCIALG_2:11
.= ((((x,(y\x)) to_power n),(x\y)) to_power (m+1)) \ (x\y) by BCIALG_2:4
.= (((x,(x\y)) to_power (m+1)),(y\x)) to_power n \ (x\y) by BCIALG_2:11;
hence thesis;
end;
theorem Th10:
Polynom (m,n+1,x,y) = Polynom (m,n,x,y) \ (y\x)
proof
consider f such that
A1: ((x,(x\y)) to_power (m+1),(y\x)) to_power (n+1) = f.(n+1) and
A2: f.0 = (x,(x\y)) to_power (m+1) and
A3: for k st k < n+1 holds f.(k + 1) = f.k \ (y\x) by BCIALG_2:def 1;
consider g such that
A4: ((x,(x\y)) to_power (m+1),(y\x)) to_power n = g.n and
A5: g.0 = (x,(x\y)) to_power (m+1) and
A6: for k st k < n holds g.(k + 1) = g.k \ (y\x) by BCIALG_2:def 1;
defpred P[Nat] means $1 <= n implies f.$1=g.$1;
now
let k;
assume
A7: k<=n implies f.k=g.k;
set m=k+1;
assume
A8: m<=n;
then k +1 <= n+1 by NAT_1:13;
then k quasi-commutative;
coherence
proof
for x,y being Element of BCI-EXAMPLE holds Polynom (0,0,x,y) = Polynom
(0,0,y,x) by STRUCT_0:def 10;
hence thesis;
end;
end;
registration
cluster quasi-commutative for BCI-algebra;
existence
proof
take BCI-EXAMPLE;
thus thesis;
end;
end;
definition
let i,j,m,n be Nat;
mode BCI-algebra of i,j,m,n -> BCI-algebra means
:Def3:
for x,y being Element of it holds Polynom (i,j,x,y) = Polynom (m,n,y,x);
existence
proof
for x,y being Element of BCI-EXAMPLE holds Polynom (i,j,x,y) = Polynom
(m,n,y,x) by STRUCT_0:def 10;
hence thesis;
end;
end;
theorem Th13:
X is BCI-algebra of i,j,m,n iff X is BCI-algebra of m,n,i,j
proof
thus X is BCI-algebra of i,j,m,n implies X is BCI-algebra of m,n,i,j
proof
assume X is BCI-algebra of i,j,m,n;
then
for x,y being Element of X holds Polynom (m,n,x,y) = Polynom (i,j,y,x)
by Def3;
hence thesis by Def3;
end;
assume X is BCI-algebra of m,n,i,j;
then for x,y being Element of X holds Polynom (m,n,y,x) = Polynom (i,j,x,y)
by Def3;
hence thesis by Def3;
end;
theorem Th14:
for X being BCI-algebra of i,j,m,n holds for k being Element of
NAT holds X is BCI-algebra of i+k,j,m,n+k
proof
let X be BCI-algebra of i,j,m,n;
let k be Element of NAT;
for x,y being Element of X holds Polynom (i+k,j,x,y) = Polynom (m,n+k,y, x)
proof
let x,y be Element of X;
A1: (Polynom (m,n,y,x),(x\y)) to_power k = Polynom (m,n+k,y,x) by BCIALG_2:10;
(Polynom (i,j,x,y),(x\y)) to_power k = ((((x,(x\y)) to_power (i+1)),(x\
y)) to_power k,(y\x)) to_power j by BCIALG_2:11
.= ((x,(x\y)) to_power (i+1+k),(y\x)) to_power j by BCIALG_2:10
.= Polynom (i+k,j,x,y);
hence thesis by A1,Def3;
end;
hence thesis by Def3;
end;
theorem
for X being BCI-algebra of i,j,m,n holds for k being Element of NAT
holds X is BCI-algebra of i,j+k,m+k,n
proof
let X be BCI-algebra of i,j,m,n;
let k be Element of NAT;
for x,y being Element of X holds Polynom (i,j+k,x,y) = Polynom (m+k,n,y, x)
proof
let x,y be Element of X;
A1: (Polynom (m,n,y,x),(y\x)) to_power k = ((((y,(y\x)) to_power (m+1)),(y
\x)) to_power k,(x\y)) to_power n by BCIALG_2:11
.= (((y,(y\x)) to_power (m+1+k)),(x\y)) to_power n by BCIALG_2:10
.= Polynom (m+k,n,y,x);
(Polynom (i,j,x,y),(y\x)) to_power k = Polynom (i,j+k,x,y) by BCIALG_2:10;
hence thesis by A1,Def3;
end;
hence thesis by Def3;
end;
registration
cluster quasi-commutative for BCK-algebra;
existence
proof
take BCI-EXAMPLE;
thus thesis;
end;
end;
registration
let i,j,m,n be Nat;
cluster being_BCK-5 for BCI-algebra of i,j,m,n;
existence
proof
for x,y being Element of BCI-EXAMPLE holds Polynom (i,j,x,y) = Polynom
(m,n,y,x) by STRUCT_0:def 10;
then reconsider B = BCI-EXAMPLE as BCI-algebra of i,j,m,n by Def3;
take B;
thus thesis;
end;
end;
definition
let i,j,m,n be Nat;
mode BCK-algebra of i,j,m,n is being_BCK-5 BCI-algebra of i,j,m,n;
end;
theorem
X is BCK-algebra of i,j,m,n iff X is BCK-algebra of m,n,i,j
proof
thus X is BCK-algebra of i,j,m,n implies X is BCK-algebra of m,n,i,j
proof
assume
A1: X is BCK-algebra of i,j,m,n;
then
for x,y being Element of X holds Polynom (m,n,x,y) = Polynom (i,j,y,x)
by Def3;
hence thesis by A1,Def3;
end;
assume
A2: X is BCK-algebra of m,n,i,j;
then for x,y being Element of X holds Polynom (m,n,y,x) = Polynom (i,j,x,y)
by Def3;
hence thesis by A2,Def3;
end;
theorem Th17:
for X being BCK-algebra of i,j,m,n holds for k being Element of
NAT holds X is BCK-algebra of i+k,j,m,n+k
proof
let X be BCK-algebra of i,j,m,n;
let k be Element of NAT;
for x,y being Element of X holds Polynom (i+k,j,x,y) = Polynom (m,n+k,y, x)
proof
let x,y be Element of X;
A1: (Polynom (m,n,y,x),(x\y)) to_power k = Polynom (m,n+k,y,x) by BCIALG_2:10;
(Polynom (i,j,x,y),(x\y)) to_power k = ((((x,(x\y)) to_power (i+1)),(x\
y)) to_power k,(y\x)) to_power j by BCIALG_2:11
.= ((x,(x\y)) to_power (i+1+k),(y\x)) to_power j by BCIALG_2:10
.= Polynom (i+k,j,x,y);
hence thesis by A1,Def3;
end;
hence thesis by Def3;
end;
theorem Th18:
for X being BCK-algebra of i,j,m,n holds for k being Element of
NAT holds X is BCK-algebra of i,j+k,m+k,n
proof
let X be BCK-algebra of i,j,m,n;
let k be Element of NAT;
for x,y being Element of X holds Polynom (i,j+k,x,y) = Polynom (m+k,n,y, x)
proof
let x,y be Element of X;
A1: (Polynom (m,n,y,x),(y\x)) to_power k = ((((y,(y\x)) to_power (m+1)),(y
\x)) to_power k,(x\y)) to_power n by BCIALG_2:11
.= (((y,(y\x)) to_power (m+1+k)),(x\y)) to_power n by BCIALG_2:10
.= Polynom (m+k,n,y,x);
(Polynom (i,j,x,y),(y\x)) to_power k = Polynom (i,j+k,x,y) by BCIALG_2:10;
hence thesis by A1,Def3;
end;
hence thesis by Def3;
end;
theorem Th19:
for X being BCK-algebra of i,j,m,n holds for x,y being Element
of X holds (x,y) to_power (i+1) = (x,y) to_power (n+1)
proof
let X be BCK-algebra of i,j,m,n;
let x,y be Element of X;
A1: (x\y)\x = (x\x)\y by BCIALG_1:7
.= y` by BCIALG_1:def 5
.= 0.X by BCIALG_1:def 8;
then
A2: Polynom (m+1,n,(x\y),x) = ((x\y),(x\(x\y))) to_power n by BCIALG_2:5
.= ((x\(x\(x\y))),(x\(x\y))) to_power n by BCIALG_1:8
.= (((x,(x\(x\y))) to_power 1),(x\(x\y))) to_power n by BCIALG_2:2
.= (x,(x\(x\y))) to_power (n+1) by BCIALG_2:10
.= (x,y) to_power (n+1) by BCIALG_2:8;
X is BCI-algebra of m,n,i,j by Th13;
then
A3: X is BCI-algebra of m+1,n,i,j+1 by Th14;
Polynom (i,j+1,x,(x\y)) = (x,(x\(x\y))) to_power (i+1) by A1,BCIALG_2:5
.= (x,y) to_power (i+1) by BCIALG_2:8;
hence thesis by A2,A3,Def3;
end;
theorem Th20:
for X being BCK-algebra of i,j,m,n holds for x,y being Element
of X holds (x,y) to_power (j+1) = (x,y) to_power (m+1)
proof
let X be BCK-algebra of i,j,m,n;
let x,y be Element of X;
A1: (x\y)\x = (x\x)\y by BCIALG_1:7
.= y` by BCIALG_1:def 5
.= 0.X by BCIALG_1:def 8;
then
A2: Polynom (i+1,j,(x\y),x) = ((x\y),(x\(x\y))) to_power j by BCIALG_2:5
.= ((x\(x\(x\y))),(x\(x\y))) to_power j by BCIALG_1:8
.= (((x,(x\(x\y))) to_power 1),(x\(x\y))) to_power j by BCIALG_2:2
.= (x,(x\(x\y))) to_power (j+1) by BCIALG_2:10
.= (x,y) to_power (j+1) by BCIALG_2:8;
A3: X is BCI-algebra of i+1,j,m,n+1 by Th14;
Polynom (m,n+1,x,(x\y)) = (x,(x\(x\y))) to_power (m+1) by A1,BCIALG_2:5
.= (x,y) to_power (m+1) by BCIALG_2:8;
hence thesis by A2,A3,Def3;
end;
theorem Th21:
for X being BCK-algebra of i,j,m,n holds X is BCK-algebra of i,j ,j,n
proof
let X be BCK-algebra of i,j,m,n;
for x,y being Element of X holds Polynom (i,j,x,y) = Polynom (j,n,y,x)
proof
let x,y be Element of X;
Polynom (i,j,x,y) = Polynom (m,n,y,x) by Def3;
hence thesis by Th20;
end;
hence thesis by Def3;
end;
theorem Th22:
for X being BCK-algebra of i,j,m,n holds X is BCK-algebra of n,j ,m,n
proof
let X be BCK-algebra of i,j,m,n;
for x,y being Element of X holds Polynom (n,j,x,y) = Polynom (m,n,y,x)
proof
let x,y be Element of X;
Polynom (i,j,x,y) = Polynom (m,n,y,x) by Def3;
hence thesis by Th19;
end;
hence thesis by Def3;
end;
:: The Reduction of the Type of Quasi-Commutative BCK-algebra
definition
let i,j,m,n;
func min(i,j,m,n) -> ExtReal equals
min(min(i,j),min(m,n));
correctness;
func max(i,j,m,n) -> ExtReal equals
max(max(i,j),max(m,n));
correctness;
end;
theorem
min(i,j,m,n) = i or min(i,j,m,n) = j or min(i,j,m,n) = m or min(i,j,m, n) = n
proof
A1: min(m,n) = m or min(m,n) = n by XXREAL_0:15;
min(i,j) = i or min(i,j) = j by XXREAL_0:15;
hence thesis by A1,XXREAL_0:15;
end;
theorem
max(i,j,m,n) = i or max(i,j,m,n) = j or max(i,j,m,n) = m or max(i,j,m, n) = n
proof
A1: max(m,n) = m or max(m,n) = n by XXREAL_0:16;
max(i,j) = i or max(i,j) = j by XXREAL_0:16;
hence thesis by A1,XXREAL_0:16;
end;
theorem Th25:
i = min(i,j,m,n) implies i<=j & i<=m & i<=n
proof
A1: min(m,n)<= n by XXREAL_0:17;
assume i = min(i,j,m,n);
then
A2: i<= min(i,j) & i<= min(m,n) by XXREAL_0:17;
min(i,j) <= j & min(m,n)<= m by XXREAL_0:17;
hence thesis by A2,A1,XXREAL_0:2;
end;
theorem Th26:
max(i,j,m,n) >= i & max(i,j,m,n) >= j & max(i,j,m,n) >= m & max(
i,j,m,n) >= n
proof
A1: max(m,n) >= m & max(m,n) >= n by XXREAL_0:25;
A2: max(i,j,m,n) >= max(i,j) & max(i,j,m,n) >= max(m,n) by XXREAL_0:25;
max(i,j) >= i & max(i,j) >= j by XXREAL_0:25;
hence thesis by A1,A2,XXREAL_0:2;
end;
theorem Th27:
for X being BCK-algebra of i,j,m,n st i = min(i,j,m,n) holds ( i
= j implies X is BCK-algebra of i,i,i,i )
proof
let X be BCK-algebra of i,j,m,n;
assume
A1: i = min(i,j,m,n);
assume
A2: i = j;
A3: for x,y being Element of X holds Polynom (i,i,x,y) <= Polynom (i,i,y,x)
proof
let x,y be Element of X;
i<= n by A1,Th25;
then
A4: ((y,(y\x)) to_power (i+1),(x\y)) to_power n <= ((y,(y\x)) to_power (i+
1),(x\y)) to_power i by Th5;
i<= m by A1,Th25;
then i+1 <= m+1 by XREAL_1:6;
then
A5: ((y,(y\x)) to_power (m+1),(x\y)) to_power n <= ((y,(y\x)) to_power (i+1
),(x\y)) to_power n by Th5,BCIALG_2:19;
Polynom (i,j,x,y) = Polynom (m,n,y,x) by Def3;
hence thesis by A2,A5,A4,Th1;
end;
for x,y being Element of X holds Polynom (i,i,y,x) = Polynom (i,i,x,y)
proof
let x,y be Element of X;
Polynom (i,i,x,y) <= Polynom (i,i,y,x) by A3;
then
A6: Polynom (i,i,x,y)\Polynom (i,i,y,x)=0.X;
Polynom (i,i,y,x) <= Polynom (i,i,x,y) by A3;
then Polynom (i,i,y,x)\Polynom (i,i,x,y)=0.X;
hence thesis by A6,BCIALG_1:def 7;
end;
hence thesis by Def3;
end;
theorem
for X being BCK-algebra of i,j,m,n st i = min(i,j,m,n) holds ( i < j &
i < n implies X is BCK-algebra of i,i+1,i,i+1 )
proof
let X be BCK-algebra of i,j,m,n;
assume
A1: i = min(i,j,m,n);
assume that
A2: i < j and
A3: i < n;
for x,y being Element of X holds Polynom (i,i+1,x,y) = Polynom (i,i+1,y, x)
proof
n - i is Element of NAT & n-i>i-i by A3,NAT_1:21,XREAL_1:9;
then n-i >=1 by NAT_1:14;
then
A4: n-i+i >= 1+i by XREAL_1:6;
m >= i by A1,Th25;
then
A5: m+1 >= i+1 by XREAL_1:6;
j - i is Element of NAT & j-i>i-i by A2,NAT_1:21,XREAL_1:9;
then j-i >=1 by NAT_1:14;
then
A6: j-i+i >= 1+i by XREAL_1:6;
let x,y be Element of X;
A7: i+1 < n+1 by A3,XREAL_1:6;
(y,(y\x)) to_power (i+1) = (y,(y\x)) to_power (n+1) by Th19;
then
A8: (y,(y\x)) to_power (i+1) = (y,(y\x)) to_power (m+1) by A7,A5,Th6;
A9: Polynom (i,j,x,y) = Polynom (m,n,y,x) & (((y,(y\x)) to_power (i+1)),(
x\y)) to_power (i+1) = (((y,(y\x)) to_power (i+1)),(x\y)) to_power (n+1)
by Def3,Th19;
(((x,(x\y)) to_power (i+1)),(y\x)) to_power (i+1) = (((x,(x\y))
to_power (i+1)),(y\x)) to_power (n+1) by Th19;
then (((x,(x\y)) to_power (i+1)),(y\x)) to_power (i+1) = (((x,(x\y))
to_power (i+1)),(y\x)) to_power j by A7,A6,Th6;
hence thesis by A7,A8,A9,A4,Th6;
end;
hence thesis by Def3;
end;
theorem
for X being BCK-algebra of i,j,m,n st i = min(i,j,m,n) holds ( i = n &
i = m implies X is BCK-algebra of i,i,i,i )
proof
let X be BCK-algebra of i,j,m,n;
assume
A1: i = min(i,j,m,n);
assume
A2: i = n & i = m;
then for x,y being Element of X holds Polynom (i,i,x,y) = Polynom (i,j,y,x)
by Def3;
then
A3: X is BCK-algebra of i,i,i,j by Def3;
i = min(i,i,i,j) by A1,A2;
hence thesis by A3,Th27;
end;
theorem
for X being BCK-algebra of i,j,m,n st i = n & m < j holds X is
BCK-algebra of i,m+1,m,i
proof
let X be BCK-algebra of i,j,m,n;
assume that
A1: i = n and
A2: m < j;
for x,y being Element of X holds Polynom (i,m+1,x,y) = Polynom (m,i,y,x)
proof
j - m is Element of NAT & j-m>m-m by A2,NAT_1:21,XREAL_1:9;
then j-m >=1 by NAT_1:14;
then
A3: j-m+m >= 1+m by XREAL_1:6;
let x,y be Element of X;
A4: m+1 < j+1 by A2,XREAL_1:6;
Polynom (i,j,x,y) = Polynom (m,n,y,x) & (((x,(x\y)) to_power (i+1)),(y
\x)) to_power (j+1) = (((x,(x\y)) to_power (i+1)),(y\x)) to_power (m+1) by Def3
,Th20;
hence thesis by A1,A4,A3,Th6;
end;
hence thesis by Def3;
end;
theorem
for X being BCK-algebra of i,j,m,n st i = n holds X is BCK-algebra of i,j,j,i
proof
let X be BCK-algebra of i,j,m,n;
assume i = n;
then reconsider X as BCK-algebra of i,j,m,i;
for x,y being Element of X holds Polynom (i,j,x,y) = Polynom (j,i,y,x)
proof
let x,y be Element of X;
Polynom (i,j,x,y) = Polynom (m,i,y,x) by Def3;
hence thesis by Th20;
end;
hence thesis by Def3;
end;
theorem
for X being BCK-algebra of i,j,m,n st l >= j & k >= n holds X is
BCK-algebra of k,l,l,k
proof
let X be BCK-algebra of i,j,m,n;
assume that
A1: l >= j and
A2: k >= n;
l - j is Element of NAT by A1,NAT_1:21;
then consider t being Element of NAT such that
A3: t=l-j;
k - n is Element of NAT by A2,NAT_1:21;
then consider t1 being Element of NAT such that
A4: t1=k-n;
X is BCK-algebra of n,j,m,n by Th22;
then X is BCK-algebra of n,j,j,n by Th21;
then X is BCK-algebra of n+t1,j,j,n+t1 by Th17;
then X is BCK-algebra of k,j+t,j+t,k by A4,Th18;
hence thesis by A3;
end;
theorem
for X being BCK-algebra of i,j,m,n st k >= max(i,j,m,n) holds X is
BCK-algebra of k,k,k,k
proof
let X be BCK-algebra of i,j,m,n;
assume
A1: k >= max(i,j,m,n);
max(i,j,m,n) >= n by Th26;
then k - n is Element of NAT by A1,NAT_1:21,XXREAL_0:2;
then consider t1 being Element of NAT such that
A2: t1=k-n;
max(i,j,m,n) >= j by Th26;
then k - j is Element of NAT by A1,NAT_1:21,XXREAL_0:2;
then consider t being Element of NAT such that
A3: t=k-j;
X is BCK-algebra of n,j,m,n by Th22;
then X is BCK-algebra of n,j,j,n by Th21;
then X is BCK-algebra of n+t1,j,j,n+t1 by Th17;
then X is BCK-algebra of k,j+t,j+t,k by A2,Th18;
hence thesis by A3;
end;
theorem
for X being BCK-algebra of i,j,m,n st i <= m & j <= n holds X is
BCK-algebra of i,j,i,j
proof
let X be BCK-algebra of i,j,m,n;
assume that
A1: i <= m and
A2: j <= n;
A3: for x,y being Element of X holds Polynom (i,j,x,y) <= Polynom (i,j,y,x)
proof
let x,y be Element of X;
i+1 <= m+1 by A1,XREAL_1:6;
then
A4: ((y,(y\x)) to_power (m+1),(x\y)) to_power n <= ((y,(y\x)) to_power (i+1
),(x\y)) to_power n by Th5,BCIALG_2:19;
Polynom (i,j,x,y) = Polynom (m,n,y,x) & ((y,(y\x)) to_power (i+1),(x\y
)) to_power n <= ((y,(y\x)) to_power (i+1),(x\y)) to_power j by A2,Def3,Th5;
hence thesis by A4,Th1;
end;
for x,y being Element of X holds Polynom (i,j,y,x) = Polynom (i,j,x,y)
proof
let x,y be Element of X;
Polynom (i,j,x,y) <= Polynom (i,j,y,x) by A3;
then
A5: Polynom (i,j,x,y)\Polynom (i,j,y,x)=0.X;
Polynom (i,j,y,x) <= Polynom (i,j,x,y) by A3;
then Polynom (i,j,y,x)\Polynom (i,j,x,y)=0.X;
hence thesis by A5,BCIALG_1:def 7;
end;
hence thesis by Def3;
end;
theorem
for X being BCK-algebra of i,j,m,n holds ( i <= m & i < n implies X is
BCK-algebra of i,j,i,i+1 )
proof
let X be BCK-algebra of i,j,m,n;
assume that
A1: i <= m and
A2: i < n;
for x,y being Element of X holds Polynom (i,j,x,y) = Polynom (i,i+1,y,x)
proof
n - i is Element of NAT & n-i>i-i by A2,NAT_1:21,XREAL_1:9;
then n-i >=1 by NAT_1:14;
then
A3: n-i+i >= 1+i by XREAL_1:6;
let x,y be Element of X;
A4: i+1 < n+1 by A2,XREAL_1:6;
A5: Polynom (i,j,x,y) = Polynom (m,n,y,x) & (((y,(y\x)) to_power (m+1)),(x
\y)) to_power (i+1) = (((y,(y\x)) to_power (m+1)),(x\y)) to_power (n+1) by Def3
,Th19;
(y,(y\x)) to_power (i+1) = (y,(y\x)) to_power (n+1) & m+1 >= i+1 by A1,Th19
,XREAL_1:6;
then (y,(y\x)) to_power (i+1) = (y,(y\x)) to_power (m+1) by A4,Th6;
hence thesis by A4,A5,A3,Th6;
end;
hence thesis by Def3;
end;
theorem Th36:
X is BCI-algebra of i,j,j+k,i+k implies X is BCK-algebra
proof
assume
A1: X is BCI-algebra of i,j,j+k,i+k;
for y be Element of X holds 0.X\y = 0.X
proof
let y be Element of X;
A2: (((y,y) to_power ((j+k)+1)),(0.X\y)) to_power (i+k) = (((y,y) to_power
(j+k)\y),(0.X\y)) to_power (i+k) by BCIALG_2:4
.= ((y\y,y) to_power (j+k),(0.X\y)) to_power (i+k) by BCIALG_2:7
.= ((0.X,y) to_power (j+k),(0.X\y)) to_power (i+k) by BCIALG_1:def 5
.= ((0.X,(0.X\y)) to_power (i+k),y) to_power (j+k) by BCIALG_2:11
.= (((0.X,0.X) to_power (i+k))\((0.X,y) to_power (i+k)),y) to_power (j
+k) by BCIALG_2:18
.= ((0.X\((0.X,y) to_power (i+k))),y) to_power (j+k) by BCIALG_2:6
.= ((0.X,y) to_power (j+k))\((0.X,y) to_power (i+k)) by BCIALG_2:7
.= ((0.X,y) to_power j,y) to_power k \ ((0.X,y) to_power (i+k)) by
BCIALG_2:10
.= ((0.X,y) to_power j,y) to_power k \ ((0.X,y) to_power i,y) to_power
k by BCIALG_2:10;
A3: ((0.X,y) to_power j,y) to_power k \ ((0.X,y) to_power i,y) to_power k
<= (0.X,y) to_power j \ (0.X,y) to_power i by BCIALG_2:21;
Polynom (i,j,0.X,y) = Polynom (j+k,i+k,y,0.X) by A1,Def3;
then
(((0.X,(0.X\y)) to_power (i+1)),y) to_power j = (((y,(y\0.X)) to_power
((j+k)+1)),(0.X\y)) to_power (i+k) by BCIALG_1:2;
then
(((0.X,(0.X\y)) to_power (i+1)),y) to_power j = (((y,y) to_power ((j+k
)+1)),(0.X\y)) to_power (i+k) by BCIALG_1:2;
then
A4: ((0.X,y) to_power j,(0.X\y)) to_power (i+1) = (((y,y) to_power ((j+k)+1
)),(0.X\y)) to_power (i+k) by BCIALG_2:11;
((0.X,y) to_power j,(0.X\y)) to_power (i+1) = ((0.X,(0.X\y)) to_power (
i+1),y) to_power j by BCIALG_2:11
.= (((0.X,0.X) to_power (i+1))\((0.X,y) to_power (i+1)),y) to_power j
by BCIALG_2:18
.= ((0.X\((0.X,y) to_power (i+1))),y) to_power j by BCIALG_2:6
.= ((0.X,y) to_power j)\((0.X,y) to_power (i+1)) by BCIALG_2:7;
then
((0.X,y) to_power j \ (0.X,y) to_power (i+1)) \ ((0.X,y) to_power j \
(0.X,y) to_power i) = 0.X by A4,A2,A3;
then 0.X \ ((0.X,y) to_power i \ (0.X,y) to_power (i+1)) = 0.X by
BCIALG_1:11;
then
A5: 0.X <= ((0.X,y) to_power i \ (0.X,y) to_power (i+1));
((0.X,y) to_power i \ (0.X,y) to_power (i+1)) = (0.X,y) to_power i \
((0.X,y) to_power i \ y) by BCIALG_2:4
.= (0.X,y) to_power i \ (0.X\y,y) to_power i by BCIALG_2:7;
then (0.X,y) to_power i \ (0.X,y) to_power (i+1) <= 0.X \ (0.X\y) by
BCIALG_2:21;
then 0.X <= 0.X \ (0.X\y) by A5,Th1;
then 0.X \ (0.X \ (0.X\y)) = 0.X;
hence thesis by BCIALG_1:8;
end;
then for x being Element of X holds x`=0.X;
hence thesis by BCIALG_1:def 8;
end;
::Some Special Classes of Quasi-Commutative BCI-algebra
theorem Th37:
X is BCI-algebra of 0,0,0,0 iff X is BCK-algebra of 0,0,0,0
proof
thus X is BCI-algebra of 0,0,0,0 implies X is BCK-algebra of 0,0,0,0
proof
assume
A1: X is BCI-algebra of 0,0,0,0;
then X is BCI-algebra of 0,0,0+0,0+0;
then reconsider X as BCK-algebra by Th36;
for x,y being Element of X holds Polynom (0,0,x,y) = Polynom (0,0,y,x)
by A1,Def3;
hence thesis by Def3;
end;
thus thesis;
end;
theorem Th38:
X is commutative BCK-algebra iff X is BCI-algebra of 0,0,0,0
proof
thus X is commutative BCK-algebra implies X is BCI-algebra of 0,0,0,0
proof
assume
A1: X is commutative BCK-algebra;
for x,y being Element of X holds Polynom (0,0,x,y) = Polynom (0,0,y,x)
proof
let x,y be Element of X;
A2: x\(x\y) = y\(y\x) by A1,BCIALG_3:def 1;
((x,(x\y)) to_power 1,(y\x)) to_power 0 = (x,(x\y)) to_power 1 by
BCIALG_2:1
.= y\(y\x) by A2,BCIALG_2:2
.= (y,(y\x)) to_power 1 by BCIALG_2:2
.= ((y,(y\x)) to_power 1,(x\y)) to_power 0 by BCIALG_2:1;
hence thesis;
end;
hence thesis by Def3;
end;
assume
A3: X is BCI-algebra of 0,0,0,0;
for x,y being Element of X holds x\(x\y) = y\(y\x)
proof
let x,y be Element of X;
A4: Polynom (0,0,x,y) = Polynom (0,0,y,x) by A3,Def3;
x\(x\y) = (x,(x\y)) to_power 1 by BCIALG_2:2
.= ((y,(y\x)) to_power 1,(x\y)) to_power 0 by A4,BCIALG_2:1
.= (y,(y\x)) to_power 1 by BCIALG_2:1
.= y\(y\x) by BCIALG_2:2;
hence thesis;
end;
hence thesis by A3,Th37,BCIALG_3:def 1;
end;
notation
let X be BCI-algebra;
synonym p-Semisimple-part(X) for AtomSet(X);
end;
reserve B,P for non empty Subset of X;
theorem
for X being BCI-algebra st B = BCK-part(X) & P = p-Semisimple-part(X)
holds B /\ P = {0.X}
proof
let X be BCI-algebra;
assume that
A1: B = BCK-part(X) and
A2: P = p-Semisimple-part(X);
thus B /\ P c= {0.X}
proof
let x be object;
assume
A3: x in B /\ P;
then
A4: x in P by XBOOLE_0:def 4;
A5: x in B by A3,XBOOLE_0:def 4;
then
A6: ex x1 being Element of X st x=x1 & 0.X<=x1 by A1;
reconsider x as Element of X by A1,A5;
A7: 0.X\x=0.X by A6;
ex x2 being Element of X st x=x2 & x2 is minimal by A2,A4;
then x=0.X by A7;
hence thesis by TARSKI:def 1;
end;
0.X in BCK-part(X) & 0.X in p-Semisimple-part(X) by BCIALG_1:19;
then 0.X in B /\ P by A1,A2,XBOOLE_0:def 4;
then for x being object st x in {0.X} holds x in B /\ P by TARSKI:def 1;
hence thesis;
end;
theorem
for X being BCI-algebra st P = p-Semisimple-part(X) holds (X is
BCK-algebra iff P = {0.X})
proof
let X be BCI-algebra;
assume
A1: P = p-Semisimple-part(X);
thus X is BCK-algebra implies P = {0.X}
proof
assume
A2: X is BCK-algebra;
thus P c= {0.X}
proof
let x be object;
assume
A3: x in P;
then
A4: ex x1 being Element of X st x=x1 & x1 is minimal by A1;
reconsider x as Element of X by A1,A3;
0.X\x = x` .= 0.X by A2,BCIALG_1:def 8;
then x=0.X by A4;
hence thesis by TARSKI:def 1;
end;
0.X in P by A1;
then for x being object st x in {0.X} holds x in P by TARSKI:def 1;
hence thesis;
end;
assume
A5: P = {0.X};
for x being Element of X holds 0.X\x=0.X
proof
let x be Element of X;
0.X in P by A5,TARSKI:def 1;
then 0.X\x in P by A1,BCIALG_1:33;
hence thesis by A5,TARSKI:def 1;
end;
then for x being Element of X holds x`=0.X;
hence thesis by BCIALG_1:def 8;
end;
theorem
for X being BCI-algebra st B = BCK-part(X) holds (X is p-Semisimple
BCI-algebra iff B = {0.X})
proof
let X be BCI-algebra;
assume
A1: B = BCK-part(X);
thus X is p-Semisimple BCI-algebra implies B = {0.X}
proof
assume
A2: X is p-Semisimple BCI-algebra;
thus B c= {0.X}
proof
let x be object;
assume
A3: x in B;
then
A4: ex x1 being Element of X st x=x1 & 0.X<=x1 by A1;
reconsider x as Element of X by A1,A3;
(x`)`=x by A2,BCIALG_1:def 26;
then (0.X)`=x by A4;
then x=0.X by BCIALG_1:def 5;
hence thesis by TARSKI:def 1;
end;
let x be object;
assume
A5: x in {0.X};
then reconsider x as Element of X by TARSKI:def 1;
x=0.X by A5,TARSKI:def 1;
then x`=0.X by BCIALG_1:def 5;
then 0.X <= x;
hence thesis by A1;
end;
assume
A6: B = {0.X};
for x being Element of X holds x`` = x
proof
let x be Element of X;
0.X\(x\(0.X\(0.X\x))) = (0.X,(x\(0.X\(0.X\x)))) to_power 1 by BCIALG_2:2
.= ((0.X,x) to_power 1)\((0.X,(0.X\(0.X\x))) to_power 1) by BCIALG_2:18
.= ((0.X,x) to_power 1)\((0.X,x) to_power 1) by BCIALG_2:8
.= 0.X by BCIALG_1:def 5;
then 0.X <= (x\(0.X\(0.X\x)));
then (x\(0.X\(0.X\x))) in B by A1;
then
A7: x\(0.X\(0.X\x)) = 0.X by A6,TARSKI:def 1;
(0.X\(0.X\x))\x = (0.X\x)\(0.X\x) by BCIALG_1:7
.= 0.X by BCIALG_1:def 5;
hence thesis by A7,BCIALG_1:def 7;
end;
hence thesis by BCIALG_1:54;
end;
theorem Th42:
X is p-Semisimple BCI-algebra implies X is BCI-algebra of 0,1,0, 0
proof
assume
A1: X is p-Semisimple BCI-algebra;
for x,y being Element of X holds Polynom (0,1,x,y) = Polynom (0,0,y,x)
proof
let x,y be Element of X;
((x,(x\y)) to_power 1,(y\x)) to_power 1 = (x\(x\y),(y\x)) to_power 1
by BCIALG_2:2
.= (y,(y\x)) to_power 1 by A1,BCIALG_1:def 26
.= ((y,(y\x)) to_power 1,(x\y)) to_power 0 by BCIALG_2:1;
hence thesis;
end;
hence thesis by Def3;
end;
theorem
X is p-Semisimple BCI-algebra implies X is BCI-algebra of n+j,n,m,m+j+ 1
proof
assume
A1: X is p-Semisimple BCI-algebra;
A2: for x,y being Element of X holds Polynom (n,n,x,y) = y
proof
let x,y be Element of X;
defpred P[Nat] means $1 <= n implies Polynom ($1,$1,x,y) = y;
now
let k;
assume
A3: k <= n implies Polynom (k,k,x,y) = y;
set m=k+1;
A4: Polynom (m,m,x,y) = Polynom (k,k+1,x,y)\(x\y) by Th9
.= (Polynom (k,k,x,y)\(y\x))\(x\y) by Th10;
assume m <= n;
then Polynom (m,m,x,y) = x\(x\y) by A1,A3,A4,BCIALG_1:def 26,NAT_1:13;
hence Polynom (m,m,x,y) = y by A1,BCIALG_1:def 26;
end;
then
A5: for k st P[k] holds P[k+1];
Polynom (0,0,x,y) = x\(x\y) by Th7
.= y by A1,BCIALG_1:def 26;
then
A6: P[0];
for n holds P[n] from NAT_1:sch 2(A6,A5);
hence thesis;
end;
A7: for x,y being Element of X holds Polynom (m,m+1,x,y) = x
proof
let x,y be Element of X;
defpred P[Nat] means
$1 <= m implies Polynom ($1,$1+1,x,y) = x;
now
let k;
assume
A8: k <= m implies Polynom (k,k+1,x,y) = x;
set l=k+1;
A9: Polynom (l,l+1,x,y) = Polynom (k,(k+1)+1,x,y)\(x\y) by Th9
.= (Polynom (k,k+1,x,y)\(y\x))\(x\y) by Th10;
assume l <= m;
then Polynom (l,l+1,x,y) = (x\(x\y))\(y\x) by A8,A9,BCIALG_1:7,NAT_1:13
.= y\(y\x) by A1,BCIALG_1:def 26;
hence Polynom (l,l+1,x,y) = x by A1,BCIALG_1:def 26;
end;
then
A10: for k st P[k] holds P[k+1];
Polynom (0,1,x,y) = Polynom (0,0,x,y)\(y\x) by Th10
.= (x\(x\y))\(y\x) by Th7
.= y\(y\x) by A1,BCIALG_1:def 26
.= x by A1,BCIALG_1:def 26;
then
A11: P[0];
for m holds P[m] from NAT_1:sch 2(A11,A10);
hence thesis;
end;
for x,y being Element of X holds Polynom (n+j,n,x,y) = Polynom (m,m+j+1 ,y,x)
proof
let x,y be Element of X;
defpred P[Nat] means $1 <= j implies Polynom (n+$1,n,x,y) =
Polynom (m,m+$1+1,y,x);
now
let k;
assume
A12: k <= j implies Polynom (n+k,n,x,y) = Polynom (m,m+k+1,y,x);
set l=k+1;
assume l <= j;
then Polynom (n+l,n,x,y) = Polynom (m,m+k+1,y,x)\(x\y) by A12,Th9,
NAT_1:13
.= Polynom (m,m+k+1+1,y,x) by Th10;
hence Polynom (n+l,n,x,y) = Polynom (m,m+l+1,y,x);
end;
then
A13: for k st P[k] holds P[k+1];
Polynom (n+0,n,x,y) = y by A2
.= Polynom (m,m+0+1,y,x) by A7;
then
A14: P[0];
for j holds P[j] from NAT_1:sch 2(A14,A13);
hence thesis;
end;
hence thesis by Def3;
end;
theorem
X is associative BCI-algebra implies X is BCI-algebra of 0,1,0,0 & X
is BCI-algebra of 1,0,0,0
proof
assume
A1: X is associative BCI-algebra;
for x being Element of X holds x`` = x
proof
let x be Element of X;
x` = x by A1,BCIALG_1:47;
hence thesis;
end;
then
A2: X is p-Semisimple by BCIALG_1:54;
for x,y being Element of X holds Polynom (1,0,x,y) = Polynom (0,0,y,x)
proof
let x,y be Element of X;
x\(x\y) = y by A2;
then
A3: (x\(x\y))\(x\y) = y\(y\x) by A1,BCIALG_1:48;
((x,(x\y)) to_power (1+1),(y\x)) to_power 0 = (x,(x\y)) to_power 2 by
BCIALG_2:1
.= (x\(x\y))\(x\y) by BCIALG_2:3
.= (y,(y\x)) to_power 1 by A3,BCIALG_2:2
.= ((y,(y\x)) to_power 1,(x\y)) to_power 0 by BCIALG_2:1;
hence thesis;
end;
hence thesis by A2,Def3,Th42;
end;
theorem
X is weakly-positive-implicative BCI-algebra implies X is BCI-algebra
of 0,1,1,1
proof
assume
A1: X is weakly-positive-implicative BCI-algebra;
for x,y being Element of X holds Polynom (0,1,x,y) = Polynom (1,1,y,x)
proof
let x,y be Element of X;
A2: (x\(x\y))\(y\x)=((y\(y\x))\(y\x))\(x\y) by A1,BCIALG_1:85;
((x,(x\y)) to_power 1,(y\x)) to_power 1 = (x,(x\y)) to_power 1 \ (y\x)
by BCIALG_2:2
.= (x\(x\y))\(y\x) by BCIALG_2:2
.= (((y\(y\x))\(y\x)),(x\y)) to_power 1 by A2,BCIALG_2:2
.= ((y,(y\x)) to_power 2,(x\y)) to_power 1 by BCIALG_2:3;
hence thesis;
end;
hence thesis by Def3;
end;
theorem
X is positive-implicative BCI-algebra implies X is BCI-algebra of 0,1, 1,1
proof
assume
A1: X is positive-implicative BCI-algebra;
for x,y being Element of X holds Polynom (0,1,x,y) = Polynom (1,1,y,x)
proof
let x,y be Element of X;
A2: (x\(x\y))\(y\x)=((y\(y\x))\(y\x))\(x\y) by A1,BCIALG_1:85;
((x,(x\y)) to_power 1,(y\x)) to_power 1 = (x,(x\y)) to_power 1 \ (y\x)
by BCIALG_2:2
.= (x\(x\y))\(y\x) by BCIALG_2:2
.= (((y\(y\x))\(y\x)),(x\y)) to_power 1 by A2,BCIALG_2:2
.= ((y,(y\x)) to_power 2,(x\y)) to_power 1 by BCIALG_2:3;
hence thesis;
end;
hence thesis by Def3;
end;
theorem
X is implicative BCI-algebra implies X is BCI-algebra of 0,1,0,0
proof
assume
A1: X is implicative BCI-algebra;
for x,y being Element of X holds Polynom (0,1,x,y) = Polynom (0,0,y,x)
proof
let x,y be Element of X;
A2: (x\(x\y))\(y\x)=y\(y\x) by A1,BCIALG_1:def 24;
((x,(x\y)) to_power 1,(y\x)) to_power 1 = (x\(x\y),(y\x)) to_power 1
by BCIALG_2:2
.= (x\(x\y))\(y\x) by BCIALG_2:2
.= (y,(y\x)) to_power 1 by A2,BCIALG_2:2
.= ((y,(y\x)) to_power 1,(x\y)) to_power 0 by BCIALG_2:1;
hence thesis;
end;
hence thesis by Def3;
end;
theorem
X is alternative BCI-algebra implies X is BCI-algebra of 0,1,0,0
proof
assume
A1: X is alternative BCI-algebra;
for x,y being Element of X holds Polynom (0,1,x,y) = Polynom (0,0,y,x)
proof
let x,y be Element of X;
A2: (x\(x\y))\(y\x)=y\(y\x) by A1,BCIALG_1:76;
((x,(x\y)) to_power 1,(y\x)) to_power 1 = (x\(x\y),(y\x)) to_power 1
by BCIALG_2:2
.= (x\(x\y))\(y\x) by BCIALG_2:2
.= (y,(y\x)) to_power 1 by A2,BCIALG_2:2
.= ((y,(y\x)) to_power 1,(x\y)) to_power 0 by BCIALG_2:1;
hence thesis;
end;
hence thesis by Def3;
end;
theorem Th49:
X is BCK-positive-implicative BCK-algebra iff X is BCK-algebra of 0,1,0,1
proof
thus X is BCK-positive-implicative BCK-algebra implies X is BCK-algebra of 0
,1,0,1
proof
assume
A1: X is BCK-positive-implicative BCK-algebra;
for x,y being Element of X holds Polynom (0,1,x,y) = Polynom (0,1,y,x)
proof
let x,y be Element of X;
(x\y)\(x\y) = 0.X by BCIALG_1:def 5;
then (x\(x\y))\y = 0.X by BCIALG_1:7;
then
A2: x\(x\y) <= y;
x\(x\y) = (x\(x\y))\(x\y) by A1,BCIALG_3:28;
then x\(x\y) <= y\(x\y) by A2,BCIALG_1:5;
then (x\(x\y))\(y\x) <= (y\(x\y))\(y\x) by BCIALG_1:5;
then
A3: (x\(x\y))\(y\x) <= (y\(y\x))\(x\y) by BCIALG_1:7;
(y\x)\(y\x) = 0.X by BCIALG_1:def 5;
then (y\(y\x))\x = 0.X by BCIALG_1:7;
then y\(y\x) <= x;
then
A4: (y\(y\x))\(y\x) <= x\(y\x) by BCIALG_1:5;
y\(y\x) = (y\(y\x))\(y\x) by A1,BCIALG_3:28;
then (y\(y\x))\(x\y) <= (x\(y\x))\(x\y) by A4,BCIALG_1:5;
then (y\(y\x))\(x\y) <= (x\(x\y))\(y\x) by BCIALG_1:7;
then
A5: (x\(x\y))\(y\x) = (y\(y\x))\(x\y) by A3,Th2;
((x,(x\y)) to_power 1,(y\x)) to_power 1 = (x,(x\y)) to_power 1 \ (y
\x) by BCIALG_2:2
.= (x\(x\y))\(y\x) by BCIALG_2:2
.= ((y\(y\x)),(x\y)) to_power 1 by A5,BCIALG_2:2
.= ((y,(y\x)) to_power 1,(x\y)) to_power 1 by BCIALG_2:2;
hence thesis;
end;
hence thesis by A1,Def3;
end;
assume
A6: X is BCK-algebra of 0,1,0,1;
for x,y being Element of X holds x\y = (x\y)\y
proof
let x,y be Element of X;
A7: Polynom (0,1,x,(x\y)) = Polynom (0,1,(x\y),x) by A6,Def3;
A8: (x\y)\x = (x\x)\y by BCIALG_1:7
.= y` by BCIALG_1:def 5
.= 0.X by A6,BCIALG_1:def 8;
then
A9: ((x\y)\((x\y)\x))\(x\(x\y)) = (x\y)\(x\(x\y)) by BCIALG_1:2
.= (x\(x\(x\y)))\y by BCIALG_1:7
.= (x\y)\y by BCIALG_1:8;
A10: (x\(x\(x\y)))\((x\y)\x) = (x\y)\((x\y)\x) by BCIALG_1:8
.= x\y by A8,BCIALG_1:2;
(x\(x\(x\y)))\((x\y)\x) = (x,(x\(x\y))) to_power 1 \ ((x\y)\x) by
BCIALG_2:2
.= (((x\y),((x\y)\x)) to_power 1,(x\(x\y))) to_power 1 by A7,BCIALG_2:2
.= (((x\y)\((x\y)\x)),(x\(x\y))) to_power 1 by BCIALG_2:2
.= ((x\y)\((x\y)\x))\(x\(x\y)) by BCIALG_2:2;
hence thesis by A10,A9;
end;
hence thesis by A6,BCIALG_3:28;
end;
theorem Th50:
X is BCK-implicative BCK-algebra iff X is BCK-algebra of 1,0,0,0
proof
thus X is BCK-implicative BCK-algebra implies X is BCK-algebra of 1,0,0,0
proof
assume
A1: X is BCK-implicative BCK-algebra;
for x,y being Element of X holds Polynom (1,0,x,y) = Polynom (0,0,y,x)
proof
let x,y be Element of X;
A2: (x\(x\y))\(x\y) = y\(y\x) by A1,BCIALG_3:35;
((x,(x\y)) to_power (1+1),(y\x)) to_power 0 = (x,(x\y)) to_power 2
by BCIALG_2:1
.= (x\(x\y))\(x\y) by BCIALG_2:3
.= (y,(y\x)) to_power 1 by A2,BCIALG_2:2
.= ((y,(y\x)) to_power 1,(x\y)) to_power 0 by BCIALG_2:1;
hence thesis;
end;
hence thesis by A1,Def3;
end;
assume
A3: X is BCK-algebra of 1,0,0,0;
for x,y being Element of X holds (x\(x\y))\(x\y) = y\(y\x)
proof
let x,y be Element of X;
A4: Polynom (1,0,x,y) = Polynom (0,0,y,x) by A3,Def3;
(x\(x\y))\(x\y) = (x,(x\y)) to_power 2 by BCIALG_2:3
.= ((x,(x\y)) to_power (1+1),(y\x)) to_power 0 by BCIALG_2:1
.= (y,(y\x)) to_power 1 by A4,BCIALG_2:1
.= y\(y\x) by BCIALG_2:2;
hence thesis;
end;
hence thesis by A3,BCIALG_3:35;
end;
registration
cluster BCK-implicative -> commutative for BCK-algebra;
coherence by BCIALG_3:34;
cluster BCK-implicative -> BCK-positive-implicative for BCK-algebra;
coherence by BCIALG_3:34;
end;
theorem
X is BCK-algebra of 1,0,0,0 iff X is BCK-algebra of 0,0,0,0 & X is
BCK-algebra of 0,1,0,1
proof
thus X is BCK-algebra of 1,0,0,0 implies X is BCK-algebra of 0,0,0,0 & X is
BCK-algebra of 0,1,0,1
proof
assume X is BCK-algebra of 1,0,0,0;
then X is BCK-implicative BCK-algebra by Th50;
hence thesis by Th38,Th49;
end;
assume X is BCK-algebra of 0,0,0,0 & X is BCK-algebra of 0,1,0,1;
then
X is commutative BCK-algebra & X is BCK-positive-implicative BCK-algebra
by Th38,Th49;
then X is BCK-implicative BCK-algebra by BCIALG_3:34;
hence thesis by Th50;
end;
theorem
for X being quasi-commutative BCK-algebra holds (X is BCK-algebra of 0
,1,0,1 iff for x,y being Element of X holds x\y = (x\y)\y )
by BCIALG_3:28,Th49;
theorem
for X being quasi-commutative BCK-algebra holds (X is BCK-algebra of n
,n+1,n,n+1 iff for x,y being Element of X holds (x,y) to_power (n+1) = (x,y)
to_power (n+2) )
proof
let X be quasi-commutative BCK-algebra;
thus X is BCK-algebra of n,n+1,n,n+1 implies for x,y being Element of X
holds (x,y) to_power (n+1) = (x,y) to_power (n+2)
proof
assume
A1: X is BCK-algebra of n,n+1,n,n+1;
for x,y being Element of X holds (x,y) to_power (n+1) = (x,y) to_power
(n+2)
proof
let x,y be Element of X;
A2: (x\y)\x = (x\x)\y by BCIALG_1:7
.= y` by BCIALG_1:def 5
.= 0.X by BCIALG_1:def 8;
then
A3: ((x,(x\(x\y))) to_power (n+1),((x\y)\x)) to_power (n+1) = ((x,y)
to_power (n+1),0.X) to_power (n+1) by BCIALG_2:8
.= (x,y) to_power (n+1) by BCIALG_2:5;
A4: (((x\y),((x\y)\x)) to_power (n+1),(x\(x\y))) to_power (n+1) = ((x\y)
,(x\(x\y))) to_power (n+1) by A2,BCIALG_2:5
.= (x,(x\(x\y))) to_power (n+1) \ y by BCIALG_2:7
.= (x,y) to_power (n+1) \ y by BCIALG_2:8
.= (x,y) to_power ((n+1)+1) by BCIALG_2:4;
Polynom (n,n+1,x,(x\y)) = Polynom (n,n+1,(x\y),x) by A1,Def3;
hence thesis by A3,A4;
end;
hence thesis;
end;
assume
A5: for x,y being Element of X holds (x,y) to_power (n+1) = (x,y)
to_power (n+2);
for x,y being Element of X holds Polynom (n,n+1,x,y) = Polynom (n,n+1,y, x)
proof
let x,y be Element of X;
(x\y)\(x\y) = 0.X by BCIALG_1:def 5;
then (x\(x\y))\y = 0.X by BCIALG_1:7;
then x\(x\y) <= y;
then ((x\(x\y)),(x\y)) to_power (n+1) <= (y,(x\y)) to_power (n+1) by
BCIALG_2:19;
then
((x,(x\y)) to_power 1,(x\y)) to_power (n+1) <= (y,(x\y)) to_power (n+
1) by BCIALG_2:2;
then
A6: (x,(x\y)) to_power (1+(n+1)) <= (y,(x\y)) to_power (n+1) by BCIALG_2:10;
(y\x)\(y\x) = 0.X by BCIALG_1:def 5;
then (y\(y\x))\x = 0.X by BCIALG_1:7;
then y\(y\x) <= x;
then ((y\(y\x)),(y\x)) to_power (n+1) <= (x,(y\x)) to_power (n+1) by
BCIALG_2:19;
then
((y,(y\x)) to_power 1,(y\x)) to_power (n+1) <= (x,(y\x)) to_power (n+
1) by BCIALG_2:2;
then
A7: (y,(y\x)) to_power (1+(n+1)) <= (x,(y\x)) to_power (n+1) by BCIALG_2:10;
(y,(y\x)) to_power (n+1) = (y,(y\x)) to_power (n+2) by A5;
then ((y,(y\x)) to_power (n+1),(x\y)) to_power (n+1) <= ((x,(y\x))
to_power (n+1),(x\y)) to_power (n+1) by A7,BCIALG_2:19;
then
A8: ((y,(y\x)) to_power (n+1),(x\y)) to_power (n+1) <= ((x,(x\y))
to_power (n+1),(y\x)) to_power (n+1) by BCIALG_2:11;
(x,(x\y)) to_power (n+1) = (x,(x\y)) to_power (n+2) by A5;
then ((x,(x\y)) to_power (n+1),(y\x)) to_power (n+1) <= ((y,(x\y))
to_power (n+1),(y\x)) to_power (n+1) by A6,BCIALG_2:19;
then
((x,(x\y)) to_power (n+1),(y\x)) to_power (n+1) <= ((y,(y\x)) to_power
(n+1),(x\y)) to_power (n+1) by BCIALG_2:11;
hence thesis by A8,Th2;
end;
hence thesis by Def3;
end;
theorem
X is BCI-algebra of 0,1,0,0 implies X is BCI-commutative BCI-algebra
proof
assume
A1: X is BCI-algebra of 0,1,0,0;
for x,y being Element of X st y\x=0.X holds y = x\(x\y)
proof
let x,y be Element of X;
Polynom (0,1,x,y) = Polynom (0,0,y,x) by A1,Def3;
then
(x,(x\y)) to_power 1 \ (y\x) = ((y,(y\x)) to_power 1,(x\y)) to_power 0
by BCIALG_2:2;
then (x\(x\y)) \ (y\x) = ((y,(y\x)) to_power 1,(x\y)) to_power 0 by
BCIALG_2:2;
then
A2: (x\(x\y)) \ (y\x) = (y,(y\x)) to_power 1 by BCIALG_2:1;
assume y\x=0.X;
then (x\(x\y)) \ 0.X = y\0.X by A2,BCIALG_2:2;
then y = (x\(x\y)) \ 0.X by BCIALG_1:2;
hence thesis by BCIALG_1:2;
end;
then for x,y being Element of X st x\y=0.X holds x = y\(y\x);
hence thesis by BCIALG_3:def 4;
end;
theorem
X is BCI-algebra of n,0,m,m implies X is BCI-commutative BCI-algebra
proof
A1: for x,y being Element of X st x\y=0.X holds (y,(y\x)) to_power (m+1) <= (
y,(y\x)) to_power 1
proof
let x,y be Element of X;
defpred P[Nat] means
$1 <= m implies (y,(y\x)) to_power ($1+1)
<= (y,(y\x)) to_power 1;
assume
A2: x\y=0.X;
now
(0.X,y\x) to_power 1 = ((0.X,y) to_power 1)\((0.X,x) to_power 1) by
BCIALG_2:18;
then 0.X\(y\x) = ((0.X,y) to_power 1)\((0.X,x) to_power 1) by BCIALG_2:2;
then
A3: 0.X\(y\x) = (0.X\y)\((0.X,x) to_power 1) by BCIALG_2:2;
let k;
assume
A4: k<= m implies (y,(y\x)) to_power (k+1) <= (y,(y\x)) to_power 1;
((0.X\y)\(0.X\x))\(x\y) = 0.X by BCIALG_1:1;
then (0.X\y)\(0.X\x) = 0.X by A2,BCIALG_1:2;
then 0.X\(y\x) = 0.X by A3,BCIALG_2:2;
then (y\y)\(y\x) = 0.X by BCIALG_1:def 5;
then (y\(y\x))\y = 0.X by BCIALG_1:7;
then y\(y\x) <= y;
then (y\(y\x),(y\x)) to_power (k+1) <= (y,(y\x)) to_power (k+1) by
BCIALG_2:19;
then
((y,(y\x)) to_power 1,(y\x)) to_power (k+1) <= (y,(y\x)) to_power (
k+1) by BCIALG_2:2;
then
A5: (y,(y\x)) to_power ((k+1)+1) <= (y,(y\x)) to_power (k+1) by BCIALG_2:10;
set m1=k+1;
assume m1<=m;
hence (y,(y\x)) to_power (m1+1) <= (y,(y\x)) to_power 1 by A4,A5,Th1,
NAT_1:13;
end;
then
A6: for k st P[k] holds P[k+1];
(y,(y\x)) to_power (0+1) \ (y,(y\x)) to_power 1 = 0.X by BCIALG_1:def 5;
then
A7: P[0] by BCIALG_1:def 11;
for m holds P[m] from NAT_1:sch 2(A7,A6);
hence thesis;
end;
assume
A8: X is BCI-algebra of n,0,m,m;
for x,y being Element of X st x\y=0.X holds x = y\(y\x)
proof
let x,y be Element of X;
assume
A9: x\y=0.X;
Polynom (n,0,x,y) = Polynom (m,m,y,x) by A8,Def3;
then (x,(y\x)) to_power 0 = ((y,(y\x)) to_power (m+1),0.X) to_power m by A9
,BCIALG_2:5;
then
(x,(y\x)) to_power 0 = ((y,(y\x)) to_power (m+1),0.X) to_power m \ 0.
X by BCIALG_1:2;
then (x,(y\x)) to_power 0 = ((y,(y\x)) to_power (m+1),0.X) to_power (m+1)
by BCIALG_2:4;
then (x,(y\x)) to_power 0 = (y,(y\x)) to_power (m+1) by BCIALG_2:5;
then x = (y,(y\x)) to_power (m+1) by BCIALG_2:1;
then x <= (y,(y\x)) to_power 1 by A1,A9;
then
A10: x <= y\(y\x) by BCIALG_2:2;
(y\(y\x))\x = (y\x)\(y\x) by BCIALG_1:7;
then (y\(y\x))\x = 0.X by BCIALG_1:def 5;
then y\(y\x) <= x;
hence thesis by A10,Th2;
end;
hence thesis by BCIALG_3:def 4;
end;
theorem
for X being BCK-algebra of i,j,m,n holds ( j = 0 & m > 0 implies X is
BCK-algebra of 0,0,0,0 )
proof
let X be BCK-algebra of i,j,m,n;
assume that
A1: j = 0 and
A2: m > 0;
for x,y being Element of X holds Polynom (0,0,x,y) = Polynom (0,n,y,x)
proof
let x,y be Element of X;
A3: Polynom (i,j,x,y) = Polynom (m,n,y,x) by Def3;
A4: i+1 >= j+1 by A1,XREAL_1:6;
(x,(x\y)) to_power (j+1) = (x,(x\y)) to_power (m+1) & j+1 < m+1 by A1,A2
,Th20,XREAL_1:6;
then (x,(x\y)) to_power (i+1) = (x,(x\y)) to_power (0+1) by A1,A4,Th6;
hence thesis by A1,A3,Th20;
end;
then reconsider X as BCK-algebra of 0,0,0,n by Def3;
A5: for x,y being Element of X holds Polynom (0,0,x,y) <= Polynom (0,0,y,x)
proof
let x,y be Element of X;
Polynom (0,0,x,y) = Polynom (0,n,y,x) by Def3;
hence thesis by Th5;
end;
for x,y being Element of X holds Polynom (0,0,y,x) = Polynom (0,0,x,y)
proof
let x,y be Element of X;
Polynom (0,0,x,y) <= Polynom (0,0,y,x) by A5;
then
A6: Polynom (0,0,x,y)\Polynom (0,0,y,x)=0.X;
Polynom (0,0,y,x) <= Polynom (0,0,x,y) by A5;
then Polynom (0,0,y,x)\Polynom (0,0,x,y)=0.X;
hence thesis by A6,BCIALG_1:def 7;
end;
hence thesis by Def3;
end;
theorem
for X being BCK-algebra of i,j,m,n holds ( m = 0 & j > 0 implies X is
BCK-algebra of 0,1,0,1 )
proof
let X be BCK-algebra of i,j,m,n;
reconsider X as BCK-algebra of i+1,j,m,n+1 by Th17;
assume that
A1: m = 0 and
A2: j > 0;
for x,y being Element of X holds Polynom (0,1,x,y) = Polynom (0,1,y,x)
proof
let x,y be Element of X;
A3: (i+1)+1 > (m+1)+0 by A1,XREAL_1:8;
A4: (((x,(x\y)) to_power (0+1)),(y\x)) to_power (j+1) = (((x,(x\y))
to_power (0+1)),(y\x)) to_power (m+1) by Th20;
A5: j+1 > m+1 by A1,A2,XREAL_1:6;
n+1 >= m+1 & (((y,(y\x)) to_power (0+1)),(x\y)) to_power (j+1) = (((y,
(y\x)) to_power (0+1)),(x\y)) to_power (m+1) by A1,Th20,XREAL_1:6;
then
A6: (((y,(y\x)) to_power (0+1)),(x\y)) to_power (0+1) = (((y,(y\x))
to_power (0+1)),(x\y)) to_power (n+1) by A1,A5,Th6;
Polynom (i+1,j,x,y) = Polynom (m,n+1,y,x) & (x,(x\y)) to_power (j+1) =
(x,(x\ y)) to_power (m+1) by Def3,Th20;
then
(((x,(x\y)) to_power (0+1)),(y\x)) to_power j = (((y,(y\x)) to_power (0
+1)),(x\y)) to_power (n+1) by A1,A5,A3,Th6;
hence thesis by A1,A5,A6,A4,Th6,NAT_1:14;
end;
hence thesis by Def3;
end;
theorem
for X being BCK-algebra of i,j,m,n holds ( n = 0 & i <> 0 implies X is
BCK-algebra of 0,0,0,0 )
proof
let X be BCK-algebra of i,j,m,n;
assume that
A1: n = 0 and
A2: i <> 0;
for x,y being Element of X holds Polynom (0,j,x,y) = Polynom (0,0,y,x)
proof
let x,y be Element of X;
A3: Polynom (i,j,x,y) = Polynom (m,n,y,x) by Def3;
A4: m+1 >= n+1 by A1,XREAL_1:6;
(y,(y\x)) to_power (n+1) = (y,(y\x)) to_power (i+1) & n+1 < i+1 by A1,A2
,Th19,XREAL_1:6;
then (y,(y\x)) to_power (m+1) = (y,(y\x)) to_power (0+1) by A1,A4,Th6;
hence thesis by A1,A3,Th19;
end;
then reconsider X as BCK-algebra of 0,j,0,0 by Def3;
A5: for x,y being Element of X holds Polynom (0,0,y,x) <= Polynom (0,0,x,y)
proof
let x,y be Element of X;
Polynom (0,j,x,y) = Polynom (0,0,y,x) by Def3;
hence thesis by Th5;
end;
for x,y being Element of X holds Polynom (0,0,y,x) = Polynom (0,0,x,y)
proof
let x,y be Element of X;
Polynom (0,0,x,y) <= Polynom (0,0,y,x) by A5;
then
A6: Polynom (0,0,x,y)\Polynom (0,0,y,x)=0.X;
Polynom (0,0,y,x) <= Polynom (0,0,x,y) by A5;
then Polynom (0,0,y,x)\Polynom (0,0,x,y)=0.X;
hence thesis by A6,BCIALG_1:def 7;
end;
hence thesis by Def3;
end;
theorem
for X being BCK-algebra of i,j,m,n holds ( i = 0 & n <> 0 implies X is
BCK-algebra of 0,1,0,1 )
proof
let X be BCK-algebra of i,j,m,n;
reconsider X as BCK-algebra of i,j+1,m+1,n by Th18;
assume that
A1: i = 0 and
A2: n <> 0;
for x,y being Element of X holds Polynom (0,1,x,y) = Polynom (0,1,y,x)
proof
let x,y be Element of X;
A3: (m+1)+1 > (i+1)+0 by A1,XREAL_1:8;
A4: (((y,(y\x)) to_power (0+1)),(x\y)) to_power (i+1) = (((y,(y\x))
to_power (0+1)),(x\y)) to_power (n+1) by Th19;
A5: n+1 > i+1 by A1,A2,XREAL_1:6;
j+1 >= i+1 & (((x,(x\y)) to_power (0+1)),(y\x)) to_power (n+1) = (((x,
(x\y)) to_power (0+1)),(y\x)) to_power (i+1) by A1,Th19,XREAL_1:6;
then
A6: (((x,(x\y)) to_power (0+1)),(y\x)) to_power (0+1) = (((x,(x\y))
to_power (0+1)),(y\x)) to_power (j+1) by A1,A5,Th6;
Polynom (i,j+1,x,y) = Polynom (m+1,n,y,x) & (y,(y\x)) to_power (n+1) =
(y,(y \x)) to_power (i+1) by Def3,Th19;
then (((x,(x\y)) to_power (0+1)),(y\x)) to_power (j+1) = (((y,(y\x))
to_power (0+1)),(x\y)) to_power n by A1,A5,A3,Th6;
hence thesis by A1,A5,A6,A4,Th6,NAT_1:14;
end;
hence thesis by Def3;
end;