Copyright (c) 1995 Association of Mizar Users
environ
vocabulary PBOOLE, FUNCT_1, ZF_REFLE, RELAT_1, FUNCT_4, CAT_1, BOOLE, CAT_4,
FUNCOP_1, ZFMISC_1, PRALG_2, AUTALG_1, FUNCT_2, TARSKI, MATRIX_1,
PRE_CIRC, FINSET_1;
notation TARSKI, XBOOLE_0, ZFMISC_1, RELAT_1, FUNCT_1, FUNCT_2, FINSET_1,
FUNCT_4, CQC_LANG, PBOOLE, PRALG_2, AUTALG_1, PRE_CIRC;
constructors CQC_LANG, PRALG_2, AUTALG_1, PRE_CIRC, MEMBERED, XBOOLE_0;
clusters SUBSET_1, PRE_CIRC, CQC_LANG, MEMBERED, ZFMISC_1, XBOOLE_0;
requirements SUBSET, BOOLE;
definitions PBOOLE;
theorems AUTALG_1, CQC_LANG, FRAENKEL, FUNCOP_1, FUNCT_4, LATTICE4, LOPCLSET,
PBOOLE, PRALG_2, PRE_CIRC, RLVECT_3, TARSKI, ZFMISC_1, XBOOLE_0,
XBOOLE_1;
schemes MSUALG_1;
begin :: Boolean of Many Sorted Sets
reserve x, y, I for set,
A, B, X, Y for ManySortedSet of I;
definition let I, A;
func bool A -> ManySortedSet of I means :Def1:
for i be set st i in I holds it.i = bool (A.i);
existence
proof
deffunc V(set) = bool (A.$1);
thus ex X being ManySortedSet of I st
for i be set st i in I holds X.i = V(i) from MSSLambda;
end;
uniqueness
proof
let X, Y be ManySortedSet of I such that
A1: (for i be set st i in I holds X.i = bool (A.i)) and
A2: for i be set st i in I holds Y.i = bool (A.i);
now
let i be set; assume
A3: i in I;
hence X.i = bool (A.i) by A1
.= Y.i by A2,A3;
end;
hence X = Y by PBOOLE:3;
end;
end;
definition let I, A;
cluster bool A -> non-empty;
coherence
proof
let i be set such that
A1: i in I;
bool (A.i) is non empty;
hence (bool A).i is non empty by A1,Def1;
end;
end;
Lm1:
for i, I, X be set
for M be ManySortedSet of I st i in I
holds dom (M +* (i .--> X)) = I
proof
let i, I, X be set,
M be ManySortedSet of I such that
A1: i in I;
thus dom (M +* (i .--> X)) = dom M \/ dom (i .--> X) by FUNCT_4:def 1
.= I \/ dom (i .--> X) by PBOOLE:def 3
.= I \/ {i} by CQC_LANG:5
.= I by A1,ZFMISC_1:46;
end;
Lm2:
for i be set st i in I holds (bool (A \/ B)).i = bool (A.i \/ B.i)
proof
let i be set; assume
A1: i in I;
hence (bool (A \/ B)).i = bool ((A \/ B).i) by Def1
.= bool (A.i \/ B.i) by A1,PBOOLE:def 7;
end;
Lm3:
for i be set st i in I holds (bool (A /\ B)).i = bool (A.i /\ B.i)
proof
let i be set; assume
A1: i in I;
hence (bool (A /\ B)).i = bool ((A /\ B).i) by Def1
.= bool (A.i /\ B.i) by A1,PBOOLE:def 8;
end;
Lm4:
for i be set st i in I holds (bool (A \ B)).i = bool (A.i \ B.i)
proof
let i be set; assume
A1: i in I;
hence (bool (A \ B)).i = bool ((A \ B).i) by Def1
.= bool (A.i \ B.i) by A1,PBOOLE:def 9;
end;
Lm5:
for i be set st i in I holds (bool (A \+\ B)).i = bool (A.i \+\ B.i)
proof
let i be set; assume
A1: i in I;
hence (bool (A \+\ B)).i = bool ((A \+\ B).i) by Def1
.= bool (A.i \+\ B.i) by A1,PBOOLE:4;
end;
theorem Th1: :: Tarski:6
X = bool Y iff for A holds A in X iff A c= Y
proof
thus X = bool Y implies for A holds A in X iff A c= Y
proof
assume
A1: X = bool Y;
let A;
thus A in X implies A c= Y
proof
assume
A2: A in X;
let i be set;
assume i in I;
then X.i = bool (Y.i) & A.i in X.i by A1,A2,Def1,PBOOLE:def 4;
hence A.i c= Y.i;
end;
assume
A3: A c= Y;
let i be set;
assume i in I;
then X.i = bool (Y.i) & A.i c= Y.i by A1,A3,Def1,PBOOLE:def 5;
hence A.i in X.i;
end;
assume
A4: for A holds A in X iff A c= Y;
now
let i be set such that
A5: i in I;
[0]I c= Y by PBOOLE:49;
then A6: [0]I in X by A4;
for A' be set holds A' in X.i iff A' c= Y.i
proof
let A' be set;
A7: dom (i .--> A') = {i} by CQC_LANG:5;
dom ([0]I +* (i .--> A')) = I by A5,Lm1;
then reconsider K = [0]I +* (i .--> A') as ManySortedSet of I by PBOOLE:
def 3;
i in {i} by TARSKI:def 1;
then A8: K.i = (i .--> A').i by A7,FUNCT_4:14
.= A' by CQC_LANG:6;
thus A' in X.i implies A' c= Y.i
proof
assume
A9: A' in X.i;
K in X
proof
let j be set such that
A10: j in I;
now per cases;
case j = i;
hence K.j in X.j by A8,A9;
case j <> i;
then not j in dom (i .--> A') by A7,TARSKI:def 1;
then K.j = [0]I.j by FUNCT_4:12;
hence K.j in X.j by A6,A10,PBOOLE:def 4;
end;
hence K.j in X.j;
end;
then K c= Y by A4;
hence A' c= Y.i by A5,A8,PBOOLE:def 5;
end;
assume
A11: A' c= Y.i;
K c= Y
proof
let j be set such that
A12: j in I;
now per cases;
case j = i;
hence K.j c= Y.j by A8,A11;
case j <> i;
then not j in dom (i .--> A') by A7,TARSKI:def 1;
then A13: K.j = [0]I.j by FUNCT_4:12;
[0]I c= Y by PBOOLE:49;
hence K.j c= Y.j by A12,A13,PBOOLE:def 5;
end;
hence K.j c= Y.j;
end;
then K in X by A4;
hence A' in X.i by A5,A8,PBOOLE:def 4;
end;
then X.i = bool (Y.i) by ZFMISC_1:def 1;
hence X.i = (bool Y).i by A5,Def1;
end;
hence thesis by PBOOLE:3;
end;
theorem :: ZFMISC_1:1
bool [0]I = I --> {{}}
proof
now
let i be set; assume
A1: i in I;
then (bool [0]I).i = bool ([0]I.i) by Def1
.= {{}} by A1,PBOOLE:5,ZFMISC_1:1;
hence (bool [0]I).i = (I --> {{}}).i by A1,FUNCOP_1:13;
end;
hence thesis by PBOOLE:3;
end;
theorem
bool (I --> x) = I --> bool x
proof
now
let i be set; assume
A1: i in I;
hence (bool (I --> x)).i = bool ((I --> x).i) by Def1
.= bool x by A1,FUNCOP_1:13
.= (I --> bool x).i by A1,FUNCOP_1:13;
end;
hence thesis by PBOOLE:3;
end;
theorem :: ZFMISC_1:30
bool (I --> {x}) = I --> { {} , {x} }
proof
now
let i be set; assume
A1: i in I;
hence (bool (I --> {x})).i = bool ((I --> {x}).i) by Def1
.= bool {x} by A1,FUNCOP_1:13
.= { {} , {x} } by ZFMISC_1:30
.= (I --> { {} , {x} }).i by A1,FUNCOP_1:13;
end;
hence thesis by PBOOLE:3;
end;
theorem :: ZFMISC_1:76
[0]I c= A
proof
let i be set; assume i in I;
then [0]I.i = {} by PBOOLE:5;
hence thesis by XBOOLE_1:2;
end;
theorem :: ZFMISC_1:79
A c= B implies bool A c= bool B
proof
assume
A1: A c= B;
let i be set; assume
A2: i in I;
then A3:(bool A).i = bool (A.i) & (bool B).i = bool (B.i) by Def1;
A.i c= B.i by A1,A2,PBOOLE:def 5;
hence (bool A).i c= (bool B).i by A3,ZFMISC_1:79;
end;
theorem :: ZFMISC_1:81
bool A \/ bool B c= bool (A \/ B)
proof
let i be set; assume
A1: i in I;
then A2:(bool A \/ bool B).i = (bool A).i \/ (bool B).i by PBOOLE:def 7
.= bool (A.i) \/ (bool B).i by A1,Def1
.= bool (A.i) \/ bool (B.i) by A1,Def1;
(bool (A \/ B)).i = bool (A.i \/ B.i) by A1,Lm2;
hence thesis by A2,ZFMISC_1:81;
end;
theorem :: ZFMISC_1:82
bool A \/ bool B = bool (A \/ B) implies
for i be set st i in I holds A.i,B.i are_c=-comparable
proof
assume
A1: bool A \/ bool B = bool (A \/ B);
let i be set such that
A2: i in I;
bool (A.i \/ B.i) = (bool A \/ bool B).i by A1,A2,Lm2
.= (bool A).i \/ (bool B).i by A2,PBOOLE:def 7
.= (bool A).i \/ (bool (B.i)) by A2,Def1
.= (bool (A.i)) \/ (bool (B.i)) by A2,Def1;
hence thesis by ZFMISC_1:82;
end;
theorem :: ZFMISC_1:83
bool (A /\ B) = bool A /\ bool B
proof
now
let i be set; assume
A1: i in I;
hence bool (A /\ B).i = bool (A.i /\ B.i) by Lm3
.= (bool (A.i)) /\ (bool (B.i)) by ZFMISC_1:83
.= (bool (A.i)) /\ (bool B.i) by A1,Def1
.= (bool A).i /\ (bool B.i) by A1,Def1
.= (bool A /\ bool B).i by A1,PBOOLE:def 8;
end;
hence thesis by PBOOLE:3;
end;
theorem :: ZFMISC_1:84
bool (A \ B) c= (I --> {{}}) \/ (bool A \ bool B)
proof
let i be set; assume
A1: i in I;
then A2:(bool (A \ B)).i = bool (A.i \ B.i) by Lm4;
((I --> {{}}) \/ (bool A \ bool B)).i = (I --> {{}}).i \/
(bool A \ bool B).i
by A1,PBOOLE:def 7
.= {{}} \/ (bool A \ bool B).i by A1,FUNCOP_1:13
.= {{}} \/ ((bool A).i \ (bool B).i) by A1,PBOOLE:def 9
.= {{}} \/ ((bool (A.i)) \ (bool B).i) by A1,Def1
.= {{}} \/ (bool (A.i) \ bool (B.i)) by A1,Def1;
hence (bool (A \ B)).i c= ((I --> {{}}) \/ (bool A \ bool B)).i
by A2,ZFMISC_1:84;
end;
theorem :: ZFMISC_1:85
X c= A \ B iff X c= A & X misses B
proof
thus X c= A \ B implies X c= A & X misses B
proof
assume X c= A \ B;
then A1: X in bool (A \ B) by Th1;
thus X c= A
proof
let i be set; assume
A2: i in I;
then X.i in (bool (A \ B)).i by A1,PBOOLE:def 4;
then X.i in bool (A.i \ B.i) by A2,Lm4;
hence X.i c= A.i by XBOOLE_1:106;
end;
let i be set; assume
A3: i in I;
then X.i in (bool (A \ B)).i by A1,PBOOLE:def 4;
then X.i in bool (A.i \ B.i) by A3,Lm4;
hence X.i misses B.i by XBOOLE_1:106;
end;
assume
A4: X c= A & X misses B;
let i be set; assume
A5: i in I;
then X.i c= A.i & X.i misses B.i by A4,PBOOLE:def 5,def 11;
then X.i c= A.i \ B.i by XBOOLE_1:86;
hence thesis by A5,PBOOLE:def 9;
end;
theorem :: ZFMISC_1:86
bool (A \ B) \/ bool (B \ A) c= bool (A \+\ B)
proof
let i be set; assume
A1: i in I;
then A2:(bool (A \ B) \/ bool (B \ A)).i = (bool (A \ B)).i \/ (bool (B \ A)).i
by PBOOLE:def 7
.= (bool (A.i \ B.i)) \/ (bool (B \ A)).i by A1,Lm4
.= (bool (A.i \ B.i)) \/ (bool (B.i \ A.i)) by A1,Lm4;
bool (A \+\ B).i = bool (A.i \+\ B.i) by A1,Lm5;
hence (bool (A \ B) \/ bool (B \ A)).i c= (bool (A \+\ B).i)
by A2,ZFMISC_1:86;
end;
theorem :: ZFMISC_1:87
X c= A \+\ B iff X c= A \/ B & X misses A /\ B
proof
thus X c= A \+\ B implies X c= A \/ B & X misses A /\ B
proof
assume X c= A \+\ B;
then A1: X in bool (A \+\ B) by Th1;
thus X c= A \/ B
proof
let i be set; assume
A2: i in I;
then X.i in (bool (A \+\ B)).i by A1,PBOOLE:def 4;
then X.i in bool (A.i \+\ B.i) by A2,Lm5;
then X.i c= A.i \/ B.i by XBOOLE_1:107;
hence X.i c= (A \/ B).i by A2,PBOOLE:def 7;
end;
let i be set; assume
A3: i in I;
then X.i in (bool (A \+\ B)).i by A1,PBOOLE:def 4;
then X.i in bool (A.i \+\ B.i) by A3,Lm5;
then X.i misses A.i /\ B.i by XBOOLE_1:107;
hence X.i misses (A /\ B).i by A3,PBOOLE:def 8;
end;
assume
A4: X c= A \/ B & X misses A /\ B;
let i be set; assume
A5: i in I;
then X.i c= (A \/ B).i & X.i misses (A /\ B).i by A4,PBOOLE:def 5,def 11;
then X.i c= A.i \/ B.i & X.i misses A.i /\ B.i by A5,PBOOLE:def 7,def 8;
then X.i c= A.i \+\ B.i by XBOOLE_1:107;
hence thesis by A5,PBOOLE:4;
end;
canceled;
theorem :: ZFMISC_1:89
X c= A or Y c= A implies X /\ Y c= A
proof
assume
A1: X c= A or Y c= A;
per cases by A1;
suppose
A2: X c= A;
let i be set; assume
A3: i in I;
then X.i c= A.i by A2,PBOOLE:def 5;
then X.i /\ Y.i c= A.i by XBOOLE_1:108;
hence thesis by A3,PBOOLE:def 8;
suppose
A4: Y c= A;
let i be set; assume
A5: i in I;
then Y.i c= A.i by A4,PBOOLE:def 5;
then X.i /\ Y.i c= A.i by XBOOLE_1:108;
hence thesis by A5,PBOOLE:def 8;
end;
theorem :: ZFMISC_1:90
X c= A implies X \ Y c= A
proof
assume
A1: X c= A;
let i be set; assume
A2: i in I;
then X.i c= A.i by A1,PBOOLE:def 5;
then X.i \ Y.i c= A.i by XBOOLE_1:109;
hence thesis by A2,PBOOLE:def 9;
end;
theorem :: ZFMISC_1:91
X c= A & Y c= A implies X \+\ Y c= A
proof
assume
A1: X c= A & Y c= A;
let i be set; assume
A2: i in I;
then X.i c= A.i & Y.i c= A.i by A1,PBOOLE:def 5;
then X.i \+\ Y.i c= A.i by XBOOLE_1:110;
then X.i \+\ Y.i in bool (A.i);
then (X \+\ Y).i in bool (A.i) by A2,PBOOLE:4;
hence thesis;
end;
theorem :: ZFMISC_1:105
[|X, Y|] c= bool bool (X \/ Y)
proof
let i be set; assume
A1: i in I;
then A2:[|X, Y|].i = [:X.i, Y.i:] by PRALG_2:def 9;
(bool bool (X \/ Y)).i = bool ((bool (X \/ Y)).i) by A1,Def1
.= bool bool (X.i \/ Y.i) by A1,Lm2;
hence [|X, Y|].i c= (bool bool (X \/ Y)).i by A2,ZFMISC_1:105;
end;
theorem :: FIN_TOPO:4
X c= A iff X in bool A
proof
thus X c= A implies X in bool A
proof
assume
A1: X c= A;
let i be set; assume
A2: i in I;
then X.i c= A.i by A1,PBOOLE:def 5;
then X.i in bool (A.i);
hence X.i in (bool A).i by A2,Def1;
end;
assume
A3: X in bool A;
let i be set; assume
A4: i in I;
then X.i in (bool A).i by A3,PBOOLE:def 4;
then X.i in bool (A.i) by A4,Def1;
hence X.i c= A.i;
end;
theorem :: FRAENKEL:5
MSFuncs (A, B) c= bool [|A, B|]
proof
let i be set; assume
A1: i in I;
then A2:(MSFuncs (A, B)).i = Funcs (A.i, B.i) by AUTALG_1:def 5;
(bool [|A, B|]).i = bool ([|A, B|].i) by A1,Def1
.= bool [:A.i, B.i:] by A1,PRALG_2:def 9;
hence (MSFuncs (A, B)).i c= (bool [|A, B|]).i by A2,FRAENKEL:5;
end;
begin :: Union of Many Sorted Sets
definition let I, A;
func union A -> ManySortedSet of I means :Def2:
for i be set st i in I holds it.i = union (A.i);
existence
proof
deffunc V(set) = union (A.$1);
thus ex X being ManySortedSet of I st
for i be set st i in I holds X.i = V(i) from MSSLambda;
end;
uniqueness
proof
let X, Y be ManySortedSet of I such that
A1: (for i be set st i in I holds X.i = union (A.i)) and
A2: for i be set st i in I holds Y.i = union (A.i);
now
let i be set; assume
A3: i in I;
hence X.i = union (A.i) by A1
.= Y.i by A2,A3;
end;
hence X = Y by PBOOLE:3;
end;
end;
definition let I;
cluster union [0]I -> empty-yielding;
coherence
proof
let i be set; assume
A1: i in I;
then union ([0]I.i) is empty by PBOOLE:5,ZFMISC_1:2;
hence (union [0]I).i is empty by A1,Def2;
end;
end;
Lm6:
for i be set st i in I holds (union (A \/ B)).i = union (A.i \/ B.i)
proof
let i be set; assume
A1: i in I;
hence (union (A \/ B)).i = union ((A \/ B).i) by Def2
.= union (A.i \/ B.i) by A1,PBOOLE:def 7;
end;
Lm7:
for i be set st i in I holds (union (A /\ B)).i = union (A.i /\ B.i)
proof
let i be set; assume
A1: i in I;
hence (union (A /\ B)).i = union ((A /\ B).i) by Def2
.= union (A.i /\ B.i) by A1,PBOOLE:def 8;
end;
theorem :: Tarski:def 4
A in union X iff ex Y st A in Y & Y in X
proof
thus A in union X implies ex Y st A in Y & Y in X
proof
assume
A1: A in union X;
defpred P[set,set] means A.$1 in $2 & $2 in X.$1;
A2: for i being set st i in I ex Y being set st P[i,Y]
proof
let i be set; assume
A3: i in I;
then A.i in (union X).i by A1,PBOOLE:def 4;
then A.i in union (X.i) by A3,Def2;
hence ex Y' be set st A.i in Y' & Y' in X.i by TARSKI:def 4;
end;
consider K be ManySortedSet of I such that
A4: for i be set st i in I holds P[i,K.i] from MSSEx(A2);
take K;
thus A in K
proof
let i be set;
assume i in I;
hence A.i in K.i by A4;
end;
thus K in X
proof
let i be set;
assume i in I;
hence K.i in X.i by A4;
end;
end;
given Y such that
A5: A in Y & Y in X;
let i be set; assume
A6: i in I;
then A.i in Y.i & Y.i in X.i by A5,PBOOLE:def 4;
then A.i in union (X.i) by TARSKI:def 4;
hence A.i in (union X.i) by A6,Def2;
end;
theorem :: ZFMISC_1:2
union [0]I = [0]I
proof
now
let i be set; assume
A1: i in I;
hence (union [0]I).i = union ([0]I.i) by Def2
.= {} by A1,PBOOLE:5,ZFMISC_1:2
.= [0]I.i by A1,PBOOLE:5;
end;
hence thesis by PBOOLE:3;
end;
theorem
union (I --> x) = I --> union x
proof
now
let i be set; assume
A1: i in I;
hence (union (I --> x)).i = union ((I --> x).i) by Def2
.= union x by A1,FUNCOP_1:13
.= (I --> union x).i by A1,FUNCOP_1:13;
end;
hence thesis by PBOOLE:3;
end;
theorem :: ZFMISC_1:31
union (I --> {x}) = I --> x
proof
now
let i be set; assume
A1: i in I;
hence (union (I --> {x})).i = union ((I --> {x}).i) by Def2
.= union {x} by A1,FUNCOP_1:13
.= x by ZFMISC_1:31
.= (I --> x).i by A1,FUNCOP_1:13;
end;
hence thesis by PBOOLE:3;
end;
theorem :: ZFMISC_1:32
union (I --> { {x},{y} }) = I --> {x,y}
proof
now
let i be set; assume
A1: i in I;
hence (union (I --> {{x},{y}})).i = union ((I --> {{x},{y}}).i) by Def2
.= union {{x},{y}} by A1,FUNCOP_1:13
.= {x,y} by ZFMISC_1:32
.= (I --> {x,y}).i by A1,FUNCOP_1:13;
end;
hence thesis by PBOOLE:3;
end;
theorem :: ZFMISC_1:92
X in A implies X c= union A
proof
assume
A1: X in A;
let i be set; assume
A2: i in I;
then X.i in A.i by A1,PBOOLE:def 4;
then X.i c= union (A.i) by ZFMISC_1:92;
hence X.i c= (union A).i by A2,Def2;
end;
theorem :: ZFMISC_1:95
A c= B implies union A c= union B
proof
assume
A1: A c= B;
let i be set; assume
A2: i in I;
then A.i c= B.i by A1,PBOOLE:def 5;
then union (A.i) c= union (B.i) by ZFMISC_1:95;
then (union A).i c= union (B.i) by A2,Def2;
hence (union A).i c= (union B).i by A2,Def2;
end;
theorem :: ZFMISC_1:96
union (A \/ B) = union A \/ union B
proof
now
let i be set; assume
A1: i in I;
hence (union (A \/ B)).i = union (A.i \/ B.i) by Lm6
.= union (A.i) \/ union (B.i) by ZFMISC_1:96
.= (union A).i \/ union (B.i) by A1,Def2
.= (union A).i \/ (union B).i by A1,Def2
.= (union A \/ union B).i by A1,PBOOLE:def 7;
end;
hence thesis by PBOOLE:3;
end;
theorem :: ZFMISC_1:97
union (A /\ B) c= union A /\ union B
proof
let i be set; assume
A1: i in I;
then A2:(union (A /\ B)).i = union (A.i /\ B.i) by Lm7;
(union A /\ union B).i = (union A).i /\ (union B).i by A1,PBOOLE:def 8
.= union (A.i) /\ (union B).i by A1,Def2
.= union (A.i) /\ union (B.i) by A1,Def2;
hence thesis by A2,ZFMISC_1:97;
end;
theorem :: ZFMISC_1:99
union bool A = A
proof
now
let i be set; assume
A1: i in I;
hence (union bool A).i = union ((bool A).i) by Def2
.= union bool (A.i) by A1,Def1
.= A.i by ZFMISC_1:99;
end;
hence thesis by PBOOLE:3;
end;
theorem :: ZFMISC_1:100
A c= bool union A
proof
let i be set; assume
A1: i in I;
then (bool union A).i = bool ((union A).i) by Def1
.= bool union (A.i) by A1,Def2;
hence thesis by ZFMISC_1:100;
end;
theorem :: LATTICE4:1
union Y c= A & X in Y implies X c= A
proof
assume
A1: union Y c= A & X in Y;
let i be set; assume
A2: i in I;
then (union Y).i c= A.i by A1,PBOOLE:def 5;
then union (Y.i) c= A.i & X.i in Y.i by A1,A2,Def2,PBOOLE:def 4;
hence X.i c= A.i by LATTICE4:1;
end;
theorem :: ZFMISC_1:94
for Z be ManySortedSet of I
for A be non-empty ManySortedSet of I holds
(for X be ManySortedSet of I st X in A holds X c= Z) implies union A c= Z
proof
let Z be ManySortedSet of I,
A be non-empty ManySortedSet of I;
assume
A1: for X be ManySortedSet of I st X in A holds X c= Z;
let i be set such that
A2: i in I;
for X' be set st X' in A.i holds X' c= Z.i
proof
let X' be set such that
A3: X' in A.i;
consider M be ManySortedSet of I such that
A4: M in A by PBOOLE:146;
A5: dom (i .--> X') = {i} by CQC_LANG:5;
dom (M +* (i .--> X')) = I by A2,Lm1;
then reconsider K = M +* (i .--> X') as ManySortedSet of I by PBOOLE:def 3;
i in {i} by TARSKI:def 1;
then A6: K.i = (i .--> X').i by A5,FUNCT_4:14
.= X' by CQC_LANG:6;
K in A
proof
let j be set such that
A7: j in I;
now per cases;
case j = i;
hence K.j in A.j by A3,A6;
case j <> i;
then not j in dom (i .--> X') by A5,TARSKI:def 1;
then K.j = M.j by FUNCT_4:12;
hence K.j in A.j by A4,A7,PBOOLE:def 4;
end;
hence K.j in A.j;
end;
then K c= Z by A1;
hence X' c= Z.i by A2,A6,PBOOLE:def 5;
end;
then union (A.i) c= Z.i by ZFMISC_1:94;
hence (union A).i c= Z.i by A2,Def2;
end;
theorem :: ZFMISC_1:98
for B be ManySortedSet of I
for A be non-empty ManySortedSet of I holds
(for X be ManySortedSet of I st X in A holds X /\ B = [0]I)
implies union(A) /\ B = [0]I
proof
let B be ManySortedSet of I,
A be non-empty ManySortedSet of I;
assume
A1: (for X be ManySortedSet of I st X in A holds X /\ B = [0]I);
now
let i be set such that
A2: i in I;
for X' be set st X' in A.i holds X' misses (B.i)
proof
let X' be set such that
A3: X' in A.i;
consider M be ManySortedSet of I such that
A4: M in A by PBOOLE:146;
A5: dom (i .--> X') = {i} by CQC_LANG:5;
dom (M +* (i .--> X')) = I by A2,Lm1;
then reconsider K = M +* (i .--> X') as ManySortedSet of I by PBOOLE:def 3;
i in {i} by TARSKI:def 1;
then A6: K.i = (i .--> X').i by A5,FUNCT_4:14
.= X' by CQC_LANG:6;
K in A
proof
let j be set such that
A7: j in I;
now per cases;
case j = i;
hence K.j in A.j by A3,A6;
case j <> i;
then not j in dom (i .--> X') by A5,TARSKI:def 1;
then K.j = M.j by FUNCT_4:12;
hence K.j in A.j by A4,A7,PBOOLE:def 4;
end;
hence K.j in A.j;
end;
then K /\ B = [0]I by A1;
then K.i /\ B.i = [0]I.i by A2,PBOOLE:def 8;
then X' /\ B.i = {} by A2,A6,PBOOLE:5;
hence X' misses B.i by XBOOLE_0:def 7;
end;
then union (A.i) misses (B.i) by ZFMISC_1:98;
then union (A.i) /\ (B.i) = {} by XBOOLE_0:def 7;
then (union A).i /\ B.i = {} by A2,Def2;
then (union A /\ B).i = {} by A2,PBOOLE:def 8;
hence (union(A) /\ B).i = [0]I.i by A2,PBOOLE:5;
end;
hence thesis by PBOOLE:3;
end;
theorem :: ZFMISC_1:101
for A, B be ManySortedSet of I st A \/ B is non-empty holds
(for X, Y be ManySortedSet of I st X <> Y & X in A \/ B & Y in A \/ B
holds X /\ Y = [0]I) implies union(A /\ B) = union A /\ union B
proof
let A, B be ManySortedSet of I such that
A1: A \/ B is non-empty;
assume
A2: for X, Y be ManySortedSet of I st X <> Y & X in A \/ B & Y in A \/ B
holds X /\ Y = [0]I;
now
let i be set such that
A3: i in I;
for X', Y' be set st X' <> Y' & X' in A.i \/ B.i & Y' in A.i \/ B.i
holds X' misses Y'
proof
let X', Y' be set such that
A4: X' <> Y' and
A5: X' in A.i \/ B.i and
A6: Y' in A.i \/ B.i;
consider M be ManySortedSet of I such that
A7: M in A \/ B by A1,PBOOLE:146;
A8: dom (i .--> X') = {i} by CQC_LANG:5;
A9: dom (M +* (i .--> X')) = I by A3,Lm1;
A10: dom (i .--> Y') = {i} by CQC_LANG:5;
dom (M +* (i .--> Y')) = I by A3,Lm1;
then reconsider Kx = M +* (i.-->X'), Ky = M +* (i.-->Y')
as ManySortedSet of I by A9,PBOOLE:def 3;
A11: i in {i} by TARSKI:def 1;
then A12: Kx.i = (i .--> X').i by A8,FUNCT_4:14
.= X' by CQC_LANG:6;
A13: Ky.i = (i .--> Y').i by A10,A11,FUNCT_4:14
.= Y' by CQC_LANG:6;
A14: Kx in A \/ B
proof
let j be set such that
A15: j in I;
now per cases;
case j = i;
hence Kx.j in (A \/ B).j by A5,A12,A15,PBOOLE:def 7;
case j <> i;
then not j in dom (i .--> X') by A8,TARSKI:def 1;
then Kx.j = M.j by FUNCT_4:12;
hence Kx.j in (A \/ B).j by A7,A15,PBOOLE:def 4;
end;
hence Kx.j in (A \/ B).j;
end;
Ky in A \/ B
proof
let j be set such that
A16: j in I;
now per cases;
case j = i;
hence Ky.j in (A \/ B).j by A6,A13,A16,PBOOLE:def 7;
case j <> i;
then not j in dom (i .--> Y') by A10,TARSKI:def 1;
then Ky.j = M.j by FUNCT_4:12;
hence Ky.j in (A \/ B).j by A7,A16,PBOOLE:def 4;
end;
hence Ky.j in (A \/ B).j;
end;
then Kx /\ Ky = [0]I by A2,A4,A12,A13,A14;
then (Kx /\ Ky).i = {} by A3,PBOOLE:5;
then X' /\ Y' = {} by A3,A12,A13,PBOOLE:def 8;
hence X' misses Y' by XBOOLE_0:def 7;
end;
then union(A.i /\ B.i) = union(A.i) /\ union(B.i) by ZFMISC_1:101;
then union(A.i /\ B.i) = (union A).i /\ union(B.i) by A3,Def2;
then union(A.i /\ B.i) = (union A).i /\ (union B).i by A3,Def2;
then union(A.i /\ B.i) = (union A /\ union B).i by A3,PBOOLE:def 8;
hence (union(A /\ B).i) = (union A /\ union B).i by A3,Lm7;
end;
hence thesis by PBOOLE:3;
end;
theorem :: LOPCLSET:31
for A, X be ManySortedSet of I
for B be non-empty ManySortedSet of I holds
(X c= union (A \/ B) & for Y be ManySortedSet of I st Y in
B holds Y /\ X = [0]I)
implies X c= union A
proof
let A, X be ManySortedSet of I,
B be non-empty ManySortedSet of I;
assume that
A1: X c= union (A \/ B) and
A2: for Y be ManySortedSet of I st Y in B holds Y /\ X = [0]I;
let i be set; assume
A3: i in I;
then X.i c= (union (A \/ B)).i by A1,PBOOLE:def 5;
then A4:X.i c= union (A.i \/ B.i) by A3,Lm6;
for Y' be set st Y' in B.i holds Y' misses X.i
proof
let Y' be set such that
A5: Y' in B.i;
consider M be ManySortedSet of I such that
A6: M in B by PBOOLE:146;
A7: dom (i .--> Y') = {i} by CQC_LANG:5;
dom (M +* (i .--> Y')) = I by A3,Lm1;
then reconsider K = M +* (i .--> Y') as ManySortedSet of I by PBOOLE:def 3;
i in {i} by TARSKI:def 1;
then A8: K.i = (i .--> Y').i by A7,FUNCT_4:14
.= Y' by CQC_LANG:6;
K in B
proof
let j be set such that
A9: j in I;
now per cases;
case j = i;
hence K.j in B.j by A5,A8;
case j <> i;
then not j in dom (i .--> Y') by A7,TARSKI:def 1;
then K.j = M.j by FUNCT_4:12;
hence K.j in B.j by A6,A9,PBOOLE:def 4;
end;
hence K.j in B.j;
end;
then K /\ X = [0]I by A2;
then (K /\ X).i = {} by A3,PBOOLE:5;
then Y' /\ X.i = {} by A3,A8,PBOOLE:def 8;
hence Y' misses X.i by XBOOLE_0:def 7;
end;
then X.i c= union (A.i) by A4,LOPCLSET:31;
hence X.i c= (union A).i by A3,Def2;
end;
theorem :: RLVECT_3:34
for A be locally-finite non-empty ManySortedSet of I st
(for X, Y be ManySortedSet of I st X in A & Y in A holds X c= Y or Y c= X)
holds union A in A
proof
let A be locally-finite non-empty ManySortedSet of I such that
A1: (for X, Y be ManySortedSet of I st X in A & Y in
A holds X c= Y or Y c= X);
let i be set; assume
A2: i in I;
then A3:A.i <> {} & A.i is finite by PBOOLE:def 16,PRE_CIRC:def 3;
for X', Y' be set st X' in A.i & Y' in A.i holds X' c= Y' or Y' c= X'
proof
let X', Y' be set such that
A4: X' in A.i & Y' in A.i;
assume
A5: not X' c= Y';
consider M be ManySortedSet of I such that
A6: M in A by PBOOLE:146;
A7: dom (i .--> Y') = {i} by CQC_LANG:5;
A8: dom (M +* (i .--> Y')) = I by A2,Lm1;
A9: dom (i .--> X') = {i} by CQC_LANG:5;
dom (M +* (i .--> X')) = I by A2,Lm1;
then reconsider K1 = M +* (i.-->X'), K2 = M +* (i.-->Y') as ManySortedSet
of I
by A8,PBOOLE:def 3;
A10: i in {i} by TARSKI:def 1;
then A11: K1.i = (i .--> X').i by A9,FUNCT_4:14
.= X' by CQC_LANG:6;
A12: K2.i = (i .--> Y').i by A7,A10,FUNCT_4:14
.= Y' by CQC_LANG:6;
A13: K1 in A
proof
let j be set such that
A14: j in I;
now per cases;
case j = i;
hence K1.j in A.j by A4,A11;
case j <> i;
then not j in dom (i .--> X') by A9,TARSKI:def 1;
then K1.j = M.j by FUNCT_4:12;
hence K1.j in A.j by A6,A14,PBOOLE:def 4;
end;
hence K1.j in A.j;
end;
K2 in A
proof
let j be set such that
A15: j in I;
now per cases;
case j = i;
hence K2.j in A.j by A4,A12;
case j <> i;
then not j in dom (i .--> Y') by A7,TARSKI:def 1;
then K2.j = M.j by FUNCT_4:12;
hence K2.j in A.j by A6,A15,PBOOLE:def 4;
end;
hence K2.j in A.j;
end;
then K1 c= K2 or K2 c= K1 by A1,A13;
hence Y' c= X' by A2,A5,A11,A12,PBOOLE:def 5;
end;
then union (A.i) in A.i by A3,RLVECT_3:34;
hence (union A).i in A.i by A2,Def2;
end;