let Y be non empty set ; :: thesis: for a, b, c, d, e being Element of Funcs Y,BOOLEAN holds
( (((a '&' b) '&' c) '&' d) '&' e '<' a & (((a '&' b) '&' c) '&' d) '&' e '<' b )
let a, b, c, d, e be Element of Funcs Y,BOOLEAN ; :: thesis: ( (((a '&' b) '&' c) '&' d) '&' e '<' a & (((a '&' b) '&' c) '&' d) '&' e '<' b )
thus (((a '&' b) '&' c) '&' d) '&' e '<' a :: thesis: (((a '&' b) '&' c) '&' d) '&' e '<' b

proof

A1: (((a '&' b) '&' c) '&' d) '&' e = (e '&' d) '&' (c '&' (b '&' a)) by BVFUNC_1:7
.= ((e '&' d) '&' c) '&' (b '&' a) by BVFUNC_1:7
.= (((e '&' d) '&' c) '&' b) '&' a by BVFUNC_1:7 ;
((((e '&' d) '&' c) '&' b) '&' a) 'imp' a = I_el Y by BVFUNC_6:39;
hence (((a '&' b) '&' c) '&' d) '&' e '<' a by A1, BVFUNC_1:19; :: thesis: verum

end;

A2: (((a '&' b) '&' c) '&' d) '&' e = (((a '&' c) '&' b) '&' d) '&' e by BVFUNC_1:7
.= (((a '&' c) '&' d) '&' b) '&' e by BVFUNC_1:7
.= (((a '&' c) '&' d) '&' e) '&' b by BVFUNC_1:7 ;
((((a '&' c) '&' d) '&' e) '&' b) 'imp' b = I_el Y by BVFUNC_6:39;
hence (((a '&' b) '&' c) '&' d) '&' e '<' b by A2, BVFUNC_1:19; :: thesis: verum