let Y be non empty set ; :: thesis: for a, b being Element of Funcs (Y,BOOLEAN) holds a 'eqv' b = (a 'or' ('not' b)) '&' (('not' a) 'or' b)
let a, b be Element of Funcs (Y,BOOLEAN); :: thesis: a 'eqv' b = (a 'or' ('not' b)) '&' (('not' a) 'or' b)
consider k3 being Function such that
A1: a 'eqv' b = k3 and
A2: dom k3 = Y and
rng k3 c= BOOLEAN by FUNCT_2:def 2;
consider k4 being Function such that
A3: (a 'or' ('not' b)) '&' (('not' a) 'or' b) = k4 and
A4: dom k4 = Y and
rng k4 c= BOOLEAN by FUNCT_2:def 2;
for x being Element of Y holds (a 'eqv' b) . x = ((a 'or' ('not' b)) '&' (('not' a) 'or' b)) . x

proof

let x be Element of Y; :: thesis: (a 'eqv' b) . x = ((a 'or' ('not' b)) '&' (('not' a) 'or' b)) . x
((a 'or' ('not' b)) '&' (('not' a) 'or' b)) . x = ((a 'or' ('not' b)) '&' (a 'imp' b)) . x by BVFUNC_4:8
.= ((a 'imp' b) '&' (b 'imp' a)) . x by BVFUNC_4:8
.= (a 'eqv' b) . x by BVFUNC_4:7 ;
hence (a 'eqv' b) . x = ((a 'or' ('not' b)) '&' (('not' a) 'or' b)) . x ; :: thesis: verum

end;

then for u being set st u in Y holds
k3 . u = k4 . u by A1, A3;
hence a 'eqv' b = (a 'or' ('not' b)) '&' (('not' a) 'or' b) by A1, A2, A3, A4, FUNCT_1:2; :: thesis: verum