:: DOMAIN_1 semantic presentation  Show TPTP formulae Show IDV graph for whole article:: Showing IDV graph ... (Click the Palm Trees again to close it)

theorem :: DOMAIN_1:1  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:2  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:3  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:4  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:5  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:6  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:7  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:8  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:9  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for a being set
for X1, X2 being non empty set st a in [:X1,X2:] holds
ex x1 being Element of X1 ex x2 being Element of X2 st a = [x1,x2]
proof end;

theorem :: DOMAIN_1:10  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:11  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:12  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1, X2 being non empty set
for x, y being Element of [:X1,X2:] st x `1 = y `1 & x `2 = y `2 holds
x = y
proof end;

definition
let X1, X2 be non empty set ;
let x1 be Element of X1;
let x2 be Element of X2;
:: original: [
redefine func [x1,x2] -> Element of [:X1,X2:];
coherence
[x1,x2] is Element of [:X1,X2:]
by ZFMISC_1:106;
end;

definition
let X1, X2 be non empty set ;
let x be Element of [:X1,X2:];
:: original: `1
redefine func x `1 -> Element of X1;
coherence
x `1 is Element of X1
by MCART_1:10;
:: original: `2
redefine func x `2 -> Element of X2;
coherence
x `2 is Element of X2
by MCART_1:10;
end;

theorem :: DOMAIN_1:13  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:14  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem Th15: :: DOMAIN_1:15  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for a being set
for X1, X2, X3 being non empty set holds
( a in [:X1,X2,X3:] iff ex x1 being Element of X1 ex x2 being Element of X2 ex x3 being Element of X3 st a = [x1,x2,x3] )
proof end;

theorem Th16: :: DOMAIN_1:16  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for D, X1, X2, X3 being non empty set st ( for a being set holds
( a in D iff ex x1 being Element of X1 ex x2 being Element of X2 ex x3 being Element of X3 st a = [x1,x2,x3] ) ) holds
D = [:X1,X2,X3:]
proof end;

theorem :: DOMAIN_1:17  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for D, X1, X2, X3 being non empty set holds
( D = [:X1,X2,X3:] iff for a being set holds
( a in D iff ex x1 being Element of X1 ex x2 being Element of X2 ex x3 being Element of X3 st a = [x1,x2,x3] ) ) by Th15, Th16;

definition
let X1, X2, X3 be non empty set ;
let x1 be Element of X1;
let x2 be Element of X2;
let x3 be Element of X3;
:: original: [
redefine func [x1,x2,x3] -> Element of [:X1,X2,X3:];
coherence
[x1,x2,x3] is Element of [:X1,X2,X3:]
by MCART_1:73;
end;

theorem :: DOMAIN_1:18  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:19  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:20  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:21  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:22  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:23  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:24  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for a being set
for X1, X2, X3 being non empty set
for x being Element of [:X1,X2,X3:] holds
( a = x `1 iff for x1 being Element of X1
for x2 being Element of X2
for x3 being Element of X3 st x = [x1,x2,x3] holds
a = x1 )
proof end;

theorem :: DOMAIN_1:25  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for b being set
for X1, X2, X3 being non empty set
for x being Element of [:X1,X2,X3:] holds
( b = x `2 iff for x1 being Element of X1
for x2 being Element of X2
for x3 being Element of X3 st x = [x1,x2,x3] holds
b = x2 )
proof end;

theorem :: DOMAIN_1:26  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for c being set
for X1, X2, X3 being non empty set
for x being Element of [:X1,X2,X3:] holds
( c = x `3 iff for x1 being Element of X1
for x2 being Element of X2
for x3 being Element of X3 st x = [x1,x2,x3] holds
c = x3 )
proof end;

theorem :: DOMAIN_1:27  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:28  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1, X2, X3 being non empty set
for x, y being Element of [:X1,X2,X3:] st x `1 = y `1 & x `2 = y `2 & x `3 = y `3 holds
x = y
proof end;

theorem :: DOMAIN_1:29  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:30  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem Th31: :: DOMAIN_1:31  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for a being set
for X1, X2, X3, X4 being non empty set holds
( a in [:X1,X2,X3,X4:] iff ex x1 being Element of X1 ex x2 being Element of X2 ex x3 being Element of X3 ex x4 being Element of X4 st a = [x1,x2,x3,x4] )
proof end;

theorem Th32: :: DOMAIN_1:32  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for D, X1, X2, X3, X4 being non empty set st ( for a being set holds
( a in D iff ex x1 being Element of X1 ex x2 being Element of X2 ex x3 being Element of X3 ex x4 being Element of X4 st a = [x1,x2,x3,x4] ) ) holds
D = [:X1,X2,X3,X4:]
proof end;

theorem :: DOMAIN_1:33  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for D, X1, X2, X3, X4 being non empty set holds
( D = [:X1,X2,X3,X4:] iff for a being set holds
( a in D iff ex x1 being Element of X1 ex x2 being Element of X2 ex x3 being Element of X3 ex x4 being Element of X4 st a = [x1,x2,x3,x4] ) ) by Th31, Th32;

definition
let X1, X2, X3, X4 be non empty set ;
let x1 be Element of X1;
let x2 be Element of X2;
let x3 be Element of X3;
let x4 be Element of X4;
:: original: [
redefine func [x1,x2,x3,x4] -> Element of [:X1,X2,X3,X4:];
coherence
[x1,x2,x3,x4] is Element of [:X1,X2,X3,X4:]
by MCART_1:84;
end;

theorem :: DOMAIN_1:34  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:35  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:36  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:37  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:38  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:39  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:40  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for a being set
for X1, X2, X3, X4 being non empty set
for x being Element of [:X1,X2,X3,X4:] holds
( a = x `1 iff for x1 being Element of X1
for x2 being Element of X2
for x3 being Element of X3
for x4 being Element of X4 st x = [x1,x2,x3,x4] holds
a = x1 )
proof end;

theorem :: DOMAIN_1:41  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for b being set
for X1, X2, X3, X4 being non empty set
for x being Element of [:X1,X2,X3,X4:] holds
( b = x `2 iff for x1 being Element of X1
for x2 being Element of X2
for x3 being Element of X3
for x4 being Element of X4 st x = [x1,x2,x3,x4] holds
b = x2 )
proof end;

theorem :: DOMAIN_1:42  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for c being set
for X1, X2, X3, X4 being non empty set
for x being Element of [:X1,X2,X3,X4:] holds
( c = x `3 iff for x1 being Element of X1
for x2 being Element of X2
for x3 being Element of X3
for x4 being Element of X4 st x = [x1,x2,x3,x4] holds
c = x3 )
proof end;

theorem :: DOMAIN_1:43  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for d being set
for X1, X2, X3, X4 being non empty set
for x being Element of [:X1,X2,X3,X4:] holds
( d = x `4 iff for x1 being Element of X1
for x2 being Element of X2
for x3 being Element of X3
for x4 being Element of X4 st x = [x1,x2,x3,x4] holds
d = x4 )
proof end;

theorem :: DOMAIN_1:44  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:45  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1, X2, X3, X4 being non empty set
for x, y being Element of [:X1,X2,X3,X4:] st x `1 = y `1 & x `2 = y `2 & x `3 = y `3 & x `4 = y `4 holds
x = y
proof end;

scheme :: DOMAIN_1:sch 1
Fraenkel1{ P1[ set ] } :
for X1 being non empty set holds { x1 where x1 is Element of X1 : P1[x1] } is Subset of X1
proof end;

scheme :: DOMAIN_1:sch 2
Fraenkel2{ P1[ set , set ] } :
for X1, X2 being non empty set holds { [x1,x2] where x1 is Element of X1, x2 is Element of X2 : P1[x1,x2] } is Subset of [:X1,X2:]
proof end;

scheme :: DOMAIN_1:sch 3
Fraenkel3{ P1[ set , set , set ] } :
for X1, X2, X3 being non empty set holds { [x1,x2,x3] where x1 is Element of X1, x2 is Element of X2, x3 is Element of X3 : P1[x1,x2,x3] } is Subset of [:X1,X2,X3:]
proof end;

scheme :: DOMAIN_1:sch 4
Fraenkel4{ P1[ set , set , set , set ] } :
for X1, X2, X3, X4 being non empty set holds { [x1,x2,x3,x4] where x1 is Element of X1, x2 is Element of X2, x3 is Element of X3, x4 is Element of X4 : P1[x1,x2,x3,x4] } is Subset of [:X1,X2,X3,X4:]
proof end;

scheme :: DOMAIN_1:sch 5
Fraenkel5{ P1[ set ], P2[ set ] } :
for X1 being non empty set st ( for x1 being Element of X1 st P1[x1] holds
P2[x1] ) holds
{ y1 where y1 is Element of X1 : P1[y1] } c= { z1 where z1 is Element of X1 : P2[z1] }
proof end;

scheme :: DOMAIN_1:sch 6
Fraenkel6{ P1[ set ], P2[ set ] } :
for X1 being non empty set st ( for x1 being Element of X1 holds
( P1[x1] iff P2[x1] ) ) holds
{ y1 where y1 is Element of X1 : P1[y1] } = { z1 where z1 is Element of X1 : P2[z1] }
proof end;

scheme :: DOMAIN_1:sch 7
SubsetD{ F1() -> non empty set , P1[ set ] } :
{ d where d is Element of F1() : P1[d] } is Subset of F1()
proof end;

theorem :: DOMAIN_1:46  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:47  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
canceled;

theorem :: DOMAIN_1:48  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1 being non empty set holds X1 = { x1 where x1 is Element of X1 : verum }
proof end;

theorem :: DOMAIN_1:49  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1, X2 being non empty set holds [:X1,X2:] = { [x1,x2] where x1 is Element of X1, x2 is Element of X2 : verum }
proof end;

theorem :: DOMAIN_1:50  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1, X2, X3 being non empty set holds [:X1,X2,X3:] = { [x1,x2,x3] where x1 is Element of X1, x2 is Element of X2, x3 is Element of X3 : verum }
proof end;

theorem :: DOMAIN_1:51  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1, X2, X3, X4 being non empty set holds [:X1,X2,X3,X4:] = { [x1,x2,x3,x4] where x1 is Element of X1, x2 is Element of X2, x3 is Element of X3, x4 is Element of X4 : verum }
proof end;

theorem :: DOMAIN_1:52  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1 being non empty set
for A1 being Subset of X1 holds A1 = { x1 where x1 is Element of X1 : x1 in A1 }
proof end;

theorem :: DOMAIN_1:53  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1, X2 being non empty set
for A1 being Subset of X1
for A2 being Subset of X2 holds [:A1,A2:] = { [x1,x2] where x1 is Element of X1, x2 is Element of X2 : ( x1 in A1 & x2 in A2 ) }
proof end;

theorem :: DOMAIN_1:54  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1, X2, X3 being non empty set
for A1 being Subset of X1
for A2 being Subset of X2
for A3 being Subset of X3 holds [:A1,A2,A3:] = { [x1,x2,x3] where x1 is Element of X1, x2 is Element of X2, x3 is Element of X3 : ( x1 in A1 & x2 in A2 & x3 in A3 ) }
proof end;

theorem :: DOMAIN_1:55  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1, X2, X3, X4 being non empty set
for A1 being Subset of X1
for A2 being Subset of X2
for A3 being Subset of X3
for A4 being Subset of X4 holds [:A1,A2,A3,A4:] = { [x1,x2,x3,x4] where x1 is Element of X1, x2 is Element of X2, x3 is Element of X3, x4 is Element of X4 : ( x1 in A1 & x2 in A2 & x3 in A3 & x4 in A4 ) }
proof end;

theorem :: DOMAIN_1:56  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1 being non empty set holds {} X1 = { x1 where x1 is Element of X1 : contradiction }
proof end;

theorem :: DOMAIN_1:57  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1 being non empty set
for A1 being Subset of X1 holds A1 ` = { x1 where x1 is Element of X1 : not x1 in A1 }
proof end;

theorem :: DOMAIN_1:58  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1 being non empty set
for A1, B1 being Subset of X1 holds A1 /\ B1 = { x1 where x1 is Element of X1 : ( x1 in A1 & x1 in B1 ) }
proof end;

theorem :: DOMAIN_1:59  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1 being non empty set
for A1, B1 being Subset of X1 holds A1 \/ B1 = { x1 where x1 is Element of X1 : ( x1 in A1 or x1 in B1 ) }
proof end;

theorem :: DOMAIN_1:60  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1 being non empty set
for A1, B1 being Subset of X1 holds A1 \ B1 = { x1 where x1 is Element of X1 : ( x1 in A1 & not x1 in B1 ) }
proof end;

theorem Th61: :: DOMAIN_1:61  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1 being non empty set
for A1, B1 being Subset of X1 holds A1 \+\ B1 = { x1 where x1 is Element of X1 : ( ( x1 in A1 & not x1 in B1 ) or ( not x1 in A1 & x1 in B1 ) ) }
proof end;

theorem Th62: :: DOMAIN_1:62  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1 being non empty set
for A1, B1 being Subset of X1 holds A1 \+\ B1 = { x1 where x1 is Element of X1 : ( not x1 in A1 iff x1 in B1 ) }
proof end;

theorem :: DOMAIN_1:63  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1 being non empty set
for A1, B1 being Subset of X1 holds A1 \+\ B1 = { x1 where x1 is Element of X1 : ( x1 in A1 iff not x1 in B1 ) }
proof end;

theorem :: DOMAIN_1:64  Show TPTP formulae Show IDV graph:: Showing IDV graph ... (Click the Palm Tree again to close it) Show TPTP problem
for X1 being non empty set
for A1, B1 being Subset of X1 holds A1 \+\ B1 = { x1 where x1 is Element of X1 : ( ( x1 in A1 & not x1 in B1 ) or ( x1 in B1 & not x1 in A1 ) ) }
proof end;

definition
let D be non empty set ;
let x1 be Element of D;
:: original: {
redefine func {x1} -> Subset of D;
coherence
{x1} is Subset of D
by SUBSET_1:55;
let x2 be Element of D;
:: original: {
redefine func {x1,x2} -> Subset of D;
coherence
{x1,x2} is Subset of D
by SUBSET_1:56;
let x3 be Element of D;
:: original: {
redefine func {x1,x2,x3} -> Subset of D;
coherence
{x1,x2,x3} is Subset of D
by SUBSET_1:57;
let x4 be Element of D;
:: original: {
redefine func {x1,x2,x3,x4} -> Subset of D;
coherence
{x1,x2,x3,x4} is Subset of D
by SUBSET_1:58;
let x5 be Element of D;
:: original: {
redefine func {x1,x2,x3,x4,x5} -> Subset of D;
coherence
{x1,x2,x3,x4,x5} is Subset of D
by SUBSET_1:59;
let x6 be Element of D;
:: original: {
redefine func {x1,x2,x3,x4,x5,x6} -> Subset of D;
coherence
{x1,x2,x3,x4,x5,x6} is Subset of D
by SUBSET_1:60;
let x7 be Element of D;
:: original: {
redefine func {x1,x2,x3,x4,x5,x6,x7} -> Subset of D;
coherence
{x1,x2,x3,x4,x5,x6,x7} is Subset of D
by SUBSET_1:61;
let x8 be Element of D;
:: original: {
redefine func {x1,x2,x3,x4,x5,x6,x7,x8} -> Subset of D;
coherence
{x1,x2,x3,x4,x5,x6,x7,x8} is Subset of D
by SUBSET_1:62;
end;

scheme :: DOMAIN_1:sch 8
SubsetFD{ F1() -> non empty set , F2() -> non empty set , F3( set ) -> Element of F2(), P1[ set ] } :
{ F3(x) where x is Element of F1() : P1[x] } is Subset of F2()
proof end;

scheme :: DOMAIN_1:sch 9
SubsetFD2{ F1() -> non empty set , F2() -> non empty set , F3() -> non empty set , F4( set , set ) -> Element of F3(), P1[ set , set ] } :
{ F4(x,y) where x is Element of F1(), y is Element of F2() : P1[x,y] } is Subset of F3()
proof end;

definition
let D1, D2, D3 be non empty set ;
let x be Element of [:[:D1,D2:],D3:];
:: original: `11
redefine func x `11 -> Element of D1;
coherence
x `11 is Element of D1
proof end;
:: original: `12
redefine func x `12 -> Element of D2;
coherence
x `12 is Element of D2
proof end;
end;

definition
let D1, D2, D3 be non empty set ;
let x be Element of [:D1,[:D2,D3:]:];
:: original: `21
redefine func x `21 -> Element of D2;
coherence
x `21 is Element of D2
proof end;
:: original: `22
redefine func x `22 -> Element of D3;
coherence
x `22 is Element of D3
proof end;
end;

scheme :: DOMAIN_1:sch 10
AndScheme{ F1() -> non empty set , P1[ set ], P2[ set ] } :
{ a where a is Element of F1() : ( P1[a] & P2[a] ) } = { a1 where a1 is Element of F1() : P1[a1] } /\ { a2 where a2 is Element of F1() : P2[a2] }
proof end;