Metamath Proof Explorer


Theorem funeq

Description: Equality theorem for function predicate. (Contributed by NM, 16-Aug-1994)

Ref Expression
Assertion funeq ⊢ A = B → Fun ⁡ A ↔ Fun ⁡ B

Proof

Step Hyp Ref Expression
1 eqimss2 ⊢ A = B → B ⊆ A
2 funss ⊢ B ⊆ A → Fun ⁡ A → Fun ⁡ B
3 1 2 syl ⊢ A = B → Fun ⁡ A → Fun ⁡ B
4 eqimss ⊢ A = B → A ⊆ B
5 funss ⊢ A ⊆ B → Fun ⁡ B → Fun ⁡ A
6 4 5 syl ⊢ A = B → Fun ⁡ B → Fun ⁡ A
7 3 6 impbid ⊢ A = B → Fun ⁡ A ↔ Fun ⁡ B