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

proof

let x be Element of Y; :: thesis: ((a' '&' b') '&' c') . x = (a' '&' (b' '&' c')) . x
A2: (a' '&' (b' '&' c')) . x = (a' . x) '&' ((b' '&' c') . x) by MARGREL1:def 21;
A3: (a' . x) '&' ((b' '&' c') . x) = (a' . x) '&' ((b' . x) '&' (c' . x)) by MARGREL1:def 21;
A4: (a' . x) '&' ((b . x) '&' (c . x)) = ((a . x) '&' (b . x)) '&' (c . x) ;
(a' . x) '&' (b' . x) = (a' '&' b') . x by MARGREL1:def 21;
hence ((a' '&' b') '&' c') . x = (a' '&' (b' '&' c')) . x by A2, A3, A4, MARGREL1:def 21; :: thesis: verum

end;

consider k3 being Function such that
A5: ( (a '&' b) '&' c = k3 & dom k3 = Y & rng k3 c= BOOLEAN ) by FUNCT_2:def 2;
consider k4 being Function such that
A6: ( a '&' (b '&' c) = k4 & dom k4 = Y & rng k4 c= BOOLEAN ) by FUNCT_2:def 2;
for u being set st u in Y holds
k3 . u = k4 . u by A1, A5, A6;
hence (a '&' b) '&' c = a '&' (b '&' c) by A5, A6, FUNCT_1:9; :: thesis: verum