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