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)
A1: 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;

consider k3 being Function such that
A2: ( a 'eqv' b = k3 & dom k3 = Y & 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 & dom k4 = Y & rng k4 c= BOOLEAN ) by FUNCT_2:def 2;
( Y = dom k3 & Y = dom k4 & ( for u being set st u in Y holds
k3 . u = k4 . u ) ) by A1, A2, A3;
hence a 'eqv' b = (a 'or' ('not' b)) '&' (('not' a) 'or' b) by A2, A3, FUNCT_1:9; :: thesis: verum