let c be 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 )
let X1, X2, X3 be 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 )
let x be Element of [:X1,X2,X3:]; ( 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 )
thus
( c = x `3 implies 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 )
( ( 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 ) implies c = x `3 )proof
A1:
x = [(x `1),(x `2),(x `3)]
by MCART_1:44;
assume A2:
c = x `3
;
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
let x1 be
Element of
X1;
for x2 being Element of X2
for x3 being Element of X3 st x = [x1,x2,x3] holds
c = x3let x2 be
Element of
X2;
for x3 being Element of X3 st x = [x1,x2,x3] holds
c = x3let x3 be
Element of
X3;
( x = [x1,x2,x3] implies c = x3 )
assume
x = [x1,x2,x3]
;
c = x3
hence
c = x3
by A2, A1, MCART_1:25;
verum
end;
thus
( ( 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 ) implies c = x `3 )
by MCART_1:67; verum