Metamath Proof Explorer


Theorem setsfun

Description: A structure with replacement is a function if the original structure is a function. (Contributed by AV, 7-Jun-2021)

Ref Expression
Assertion setsfun G V Fun G I U E W Fun G sSet I E

Proof

Step Hyp Ref Expression
1 funres Fun G Fun G V dom I E
2 1 adantl G V Fun G Fun G V dom I E
3 2 adantr G V Fun G I U E W Fun G V dom I E
4 funsng I U E W Fun I E
5 4 adantl G V Fun G I U E W Fun I E
6 dmres dom G V dom I E = V dom I E dom G
7 6 ineq1i dom G V dom I E dom I E = V dom I E dom G dom I E
8 in32 V dom I E dom G dom I E = V dom I E dom I E dom G
9 disjdifr V dom I E dom I E =
10 9 ineq1i V dom I E dom I E dom G = dom G
11 0in dom G =
12 8 10 11 3eqtri V dom I E dom G dom I E =
13 7 12 eqtri dom G V dom I E dom I E =
14 13 a1i G V Fun G I U E W dom G V dom I E dom I E =
15 funun Fun G V dom I E Fun I E dom G V dom I E dom I E = Fun G V dom I E I E
16 3 5 14 15 syl21anc G V Fun G I U E W Fun G V dom I E I E
17 opex I E V
18 17 a1i Fun G I E V
19 setsvalg G V I E V G sSet I E = G V dom I E I E
20 18 19 sylan2 G V Fun G G sSet I E = G V dom I E I E
21 20 funeqd G V Fun G Fun G sSet I E Fun G V dom I E I E
22 21 adantr G V Fun G I U E W Fun G sSet I E Fun G V dom I E I E
23 16 22 mpbird G V Fun G I U E W Fun G sSet I E