let s1, s2 be Function; :: thesis: ( dom s1 = (dom p) /\ (dom q) & ( for x being set st x in dom s1 holds
s1 . x = (p . x) 'nand' (q . x) ) & dom s2 = (dom p) /\ (dom q) & ( for x being set st x in dom s2 holds
s2 . x = (p . x) 'nand' (q . x) ) implies s1 = s2 )
assume that
A2: dom s1 = (dom p) /\ (dom q) and
A3: for x being set st x in dom s1 holds
s1 . x = (p . x) 'nand' (q . x) and
A4: dom s2 = (dom p) /\ (dom q) and
A5: for x being set st x in dom s2 holds
s2 . x = (p . x) 'nand' (q . x) ; :: thesis: s1 = s2
for x being object st x in dom s1 holds
s1 . x = s2 . x

proof

let x be object ; :: thesis: ( x in dom s1 implies s1 . x = s2 . x )
assume A6: x in dom s1 ; :: thesis: s1 . x = s2 . x
then s1 . x = (p . x) 'nand' (q . x) by A3;
hence s1 . x = s2 . x by A2, A4, A5, A6; :: thesis: verum

end;

hence s1 = s2 by A2, A4, FUNCT_1:2; :: thesis: verum