let A, B, C be set ; :: thesis: ( ( not C = {} or B = {} or A = {} ) implies for f being Function of A,(Funcs (B,C)) holds rng () c= Funcs (A,C) )
assume ( not C = {} or B = {} or A = {} ) ; :: thesis: for f being Function of A,(Funcs (B,C)) holds rng () c= Funcs (A,C)
then A1: ( Funcs (B,C) = {} implies A = {} ) by FUNCT_2:8;
let f be Function of A,(Funcs (B,C)); :: thesis: rng () c= Funcs (A,C)
A2: dom (rngs f) = dom f by FUNCT_6:def 3;
then A3: dom (rngs f) = A by ;
A4: for x being object st x in dom (rngs f) holds
(rngs f) . x c= (A --> C) . x
proof
let x be object ; :: thesis: ( x in dom (rngs f) implies (rngs f) . x c= (A --> C) . x )
assume A5: x in dom (rngs f) ; :: thesis: (rngs f) . x c= (A --> C) . x
A6: (rngs f) . x = rng (f . x) by ;
f . x in Funcs (B,C) by ;
then (rngs f) . x c= C by ;
hence (rngs f) . x c= (A --> C) . x by ; :: thesis: verum
end;
dom (rngs f) = dom (A --> C) by A3;
then product (rngs f) c= product (A --> C) by ;
then product (rngs f) c= Funcs (A,C) by CARD_3:11;
hence rng () c= Funcs (A,C) by FUNCT_6:38; :: thesis: verum