let Y be non empty set ; :: thesis: for a, b being Element of Funcs Y,BOOLEAN holds a = (a '&' b) 'or' (a '&' ('not' b))
let a, b be Element of Funcs Y,BOOLEAN ; :: thesis: a = (a '&' b) 'or' (a '&' ('not' b))
A1: for x being Element of Y holds a . x = ((a '&' b) 'or' (a '&' ('not' b))) . x

proof

let x be Element of Y; :: thesis: a . x = ((a '&' b) 'or' (a '&' ('not' b))) . x
((a '&' b) 'or' (a '&' ('not' b))) . x = (a '&' (b 'or' ('not' b))) . x by BVFUNC_1:15
.= (a '&' (I_el Y)) . x by BVFUNC_4:6
.= a . x by BVFUNC_1:9 ;
hence a . x = ((a '&' b) 'or' (a '&' ('not' b))) . x ; :: thesis: verum

end;

consider k3 being Function such that
A2: ( a = k3 & dom k3 = Y & rng k3 c= BOOLEAN ) by FUNCT_2:def 2;
consider k4 being Function such that
A3: ( (a '&' b) 'or' (a '&' ('not' 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 = (a '&' b) 'or' (a '&' ('not' b)) by A2, A3, FUNCT_1:9; :: thesis: verum