let IT be Function of A,BOOLEAN; :: thesis: ( IT = p 'eqv' q iff for x being Element of A holds IT . x = 'not' ((p . x) 'xor' (q . x)) )
A7: dom q = A by FUNCT_2:def 1;

hereby :: thesis: ( ( for x being Element of A holds IT . x = 'not' ((p . x) 'xor' (q . x)) ) implies IT = p 'eqv' q )

assume A8: IT = p 'eqv' q ; :: thesis: for x being Element of A holds IT . x = 'not' ((p . x) 'xor' (q . x))
let x be Element of A; :: thesis: IT . x = 'not' ((p . x) 'xor' (q . x))
( dom p = A & dom q = A ) by FUNCT_2:def 1;
then dom (p 'eqv' q) = A /\ A by Def7
.= A ;
hence IT . x = 'not' ((p . x) 'xor' (q . x)) by A8, Def7; :: thesis: verum

end;

assume A9: for x being Element of A holds IT . x = 'not' ((p . x) 'xor' (q . x)) ; :: thesis: IT = p 'eqv' q
A10: for x being set st x in dom IT holds
IT . x = (p . x) <=> (q . x)

proof

let x be set ; :: thesis: ( x in dom IT implies IT . x = (p . x) <=> (q . x) )
assume x in dom IT ; :: thesis: IT . x = (p . x) <=> (q . x)
then reconsider x = x as Element of A by FUNCT_2:def 1;
IT . x = 'not' ((p . x) 'xor' (q . x)) by A9;
hence IT . x = (p . x) <=> (q . x) ; :: thesis: verum

end;

dom IT = A /\ A by FUNCT_2:def 1
.= (dom p) /\ (dom q) by A7, FUNCT_2:def 1 ;
hence IT = p 'eqv' q by A10, Def7; :: thesis: verum