let Y be non empty set ; :: thesis: for a, b, c, d being Function of Y,BOOLEAN st a 'imp' b = I_el Y & c 'imp' d = I_el Y holds
(a 'or' c) 'imp' (b 'or' d) = I_el Y
let a, b, c, d be Function of Y,BOOLEAN; :: thesis: ( a 'imp' b = I_el Y & c 'imp' d = I_el Y implies (a 'or' c) 'imp' (b 'or' d) = I_el Y )
assume that
A1: a 'imp' b = I_el Y and
A2: c 'imp' d = I_el Y ; :: thesis: (a 'or' c) 'imp' (b 'or' d) = I_el Y
for x being Element of Y holds ((a 'or' c) 'imp' (b 'or' d)) . x = TRUE

proof

let x be Element of Y; :: thesis: ((a 'or' c) 'imp' (b 'or' d)) . x = TRUE
(a 'imp' b) . x = TRUE by A1, BVFUNC_1:def 11;
then A3: ('not' (a . x)) 'or' (b . x) = TRUE by BVFUNC_1:def 8;
(c 'imp' d) . x = TRUE by A2, BVFUNC_1:def 11;
then A4: ('not' (c . x)) 'or' (d . x) = TRUE by BVFUNC_1:def 8;
((a 'or' c) 'imp' (b 'or' d)) . x = ('not' ((a 'or' c) . x)) 'or' ((b 'or' d) . x) by BVFUNC_1:def 8
.= ('not' ((a . x) 'or' (c . x))) 'or' ((b 'or' d) . x) by BVFUNC_1:def 4
.= ((b . x) 'or' (d . x)) 'or' (('not' (a . x)) '&' ('not' (c . x))) by BVFUNC_1:def 4
.= (((d . x) 'or' (b . x)) 'or' ('not' (a . x))) '&' (((b . x) 'or' (d . x)) 'or' ('not' (c . x))) by XBOOLEAN:9
.= ((d . x) 'or' ((b . x) 'or' ('not' (a . x)))) '&' (((b . x) 'or' (d . x)) 'or' ('not' (c . x)))
.= TRUE '&' ((b . x) 'or' TRUE) by A3, A4, BINARITH:11
.= TRUE ;
hence ((a 'or' c) 'imp' (b 'or' d)) . x = TRUE ; :: thesis: verum

end;

hence (a 'or' c) 'imp' (b 'or' d) = I_el Y by BVFUNC_1:def 11; :: thesis: verum