let Y be non empty set ; :: thesis: for a, b being Element of Funcs (Y,BOOLEAN) st ('not' a) 'imp' b = I_el Y holds
('not' b) 'imp' a = I_el Y
let a, b be Element of Funcs (Y,BOOLEAN); :: thesis: ( ('not' a) 'imp' b = I_el Y implies ('not' b) 'imp' a = I_el Y )
assume A1: ('not' a) 'imp' b = I_el Y ; :: thesis: ('not' b) 'imp' a = I_el Y
for x being Element of Y holds (('not' b) 'imp' a) . x = TRUE

proof

let x be Element of Y; :: thesis: (('not' b) 'imp' a) . x = TRUE
(('not' a) 'imp' b) . x = TRUE by A1, BVFUNC_1:def 14;
then ('not' (('not' a) . x)) 'or' (b . x) = TRUE by BVFUNC_1:def 11;
then A2: ('not' ('not' (a . x))) 'or' (b . x) = TRUE by MARGREL1:def 20;
(('not' b) 'imp' a) . x = ('not' (('not' b) . x)) 'or' (a . x) by BVFUNC_1:def 11
.= TRUE by A2, MARGREL1:def 20 ;
hence (('not' b) 'imp' a) . x = TRUE ; :: thesis: verum

end;

hence ('not' b) 'imp' a = I_el Y by BVFUNC_1:def 14; :: thesis: verum