let Y be non empty set ; :: thesis: for a, b, c being Element of Funcs Y,BOOLEAN holds (a '&' b) 'imp' c = a 'imp' (b 'imp' c)
let a, b, c be Element of Funcs Y,BOOLEAN ; :: thesis: (a '&' b) 'imp' c = a 'imp' (b 'imp' c)
A1: for x being Element of Y holds ((a '&' b) 'imp' c) . x = (a 'imp' (b 'imp' c)) . x

proof

let x be Element of Y; :: thesis: ((a '&' b) 'imp' c) . x = (a 'imp' (b 'imp' c)) . x
(a 'imp' (b 'imp' c)) . x = ('not' (a . x)) 'or' ((b 'imp' c) . x) by BVFUNC_1:def 11
.= ('not' (a . x)) 'or' (('not' (b . x)) 'or' (c . x)) by BVFUNC_1:def 11
.= (('not' (a . x)) 'or' ('not' (b . x))) 'or' (c . x) by BINARITH:20
.= ('not' ((a '&' b) . x)) 'or' (c . x) by MARGREL1:def 21
.= ((a '&' b) 'imp' c) . x by BVFUNC_1:def 11 ;
hence ((a '&' b) 'imp' c) . x = (a 'imp' (b 'imp' c)) . x ; :: thesis: verum

end;

consider k3 being Function such that
A2: ( (a '&' b) 'imp' c = k3 & dom k3 = Y & rng k3 c= BOOLEAN ) by FUNCT_2:def 2;
consider k4 being Function such that
A3: ( a 'imp' (b 'imp' c) = 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 '&' b) 'imp' c = a 'imp' (b 'imp' c) by A2, A3, FUNCT_1:9; :: thesis: verum