let C be non empty set ; :: thesis: for f, g, h being Membership_Func of C holds f * (min g,h) = min (f * g),(f * h)
let f, g, h be Membership_Func of C; :: thesis: f * (min g,h) = min (f * g),(f * h)
A1:
( C = dom (f * (min g,h)) & C = dom (min (f * g),(f * h)) )
by FUNCT_2:def 1;
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 )
A2:
f . c >= 0
by Th1;
(f * (min g,h)) . c =
(f . c) * ((min g,h) . c)
by Def2
.=
(f . c) * (min (g . c),(h . c))
by FUZZY_1:6
.=
min ((f . c) * (g . c)),
((f . c) * (h . c))
by A2, Lm1
.=
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_1:6;
:: thesis: verum
end;
hence
f * (min g,h) = min (f * g),(f * h)
by A1, PARTFUN1:34; :: thesis: verum