let Y be non empty set ; :: thesis: for a, b being Element of Funcs Y,BOOLEAN holds (a 'imp' b) '&' (('not' a) 'imp' b) '<' b
let a, b be Element of Funcs Y,BOOLEAN ; :: thesis: (a 'imp' b) '&' (('not' a) 'imp' b) '<' b
let z be Element of Y; :: according to BVFUNC_1:def 15 :: thesis: ( not ((a 'imp' b) '&' (('not' a) 'imp' b)) . z = TRUE or b . z = TRUE )
reconsider az = a . z as boolean set ;
A1: ((a 'imp' b) '&' (('not' a) 'imp' b)) . z = ((a 'imp' b) . z) '&' ((('not' a) 'imp' b) . z) by MARGREL1:def 21
.= ((('not' a) 'or' b) . z) '&' ((('not' a) 'imp' b) . z) by BVFUNC_4:8
.= ((('not' a) 'or' b) . z) '&' ((('not' ('not' a)) 'or' b) . z) by BVFUNC_4:8
.= ((('not' a) . z) 'or' (b . z)) '&' ((a 'or' b) . z) by BVFUNC_1:def 7
.= ((('not' a) . z) 'or' (b . z)) '&' ((a . z) 'or' (b . z)) by BVFUNC_1:def 7 ;
assume A2: ((a 'imp' b) '&' (('not' a) 'imp' b)) . z = TRUE ; :: thesis: b . z = TRUE

now

assume b . z <> TRUE ; :: thesis: contradiction
then b . z = FALSE by XBOOLEAN:def 3;
then ((('not' a) . z) 'or' (b . z)) '&' ((a . z) 'or' (b . z)) = ('not' az) '&' az by MARGREL1:def 20
.= FALSE by XBOOLEAN:138 ;
hence contradiction by A2, A1; :: thesis: verum

end;

hence b . z = TRUE ; :: thesis: verum