let C be non empty set ; :: thesis: for f, h, g being Membership_Func of C holds f * (min (g,h)) = min ((f * g),(f * h))
let f, h, g be Membership_Func of C; :: thesis: f * (min (g,h)) = min ((f * g),(f * h))
A1: C = dom (min ((f * g),(f * h))) by FUNCT_2:def 1;
A2: for c being Element of C st c in C holds
(f * (min (g,h))) . c = (min ((f * g),(f * h))) . c

proof

let c be Element of C; :: thesis: ( c in C implies (f * (min (g,h))) . c = (min ((f * g),(f * h))) . c )
(f * (min (g,h))) . c = (f . c) * ((min (g,h)) . c) by Def2
.= (f . c) * (min ((g . c),(h . c))) by FUZZY_1:5
.= min (((f . c) * (g . c)),((f . c) * (h . c))) by Lm1, Th1
.= min (((f * g) . c),((f . c) * (h . c))) by Def2
.= min (((f * g) . c),((f * h) . c)) by Def2 ;
hence ( c in C implies (f * (min (g,h))) . c = (min ((f * g),(f * h))) . c ) by FUZZY_1:5; :: thesis: verum

end;

C = dom (f * (min (g,h))) by FUNCT_2:def 1;
hence f * (min (g,h)) = min ((f * g),(f * h)) by A1, A2, PARTFUN1:5; :: thesis: verum