Metamath Proof Explorer


Theorem ax9

Description: Proof of ax-9 from ax9v1 and ax9v2 , proving sufficiency of the conjunction of the latter two weakened versions of ax9v , which is itself a weakened version of ax-9 . (Contributed by BJ, 7-Dec-2020) (Proof shortened by Wolf Lammen, 11-Apr-2021)

Ref Expression
Assertion ax9 ⊢ x = y → z ∈ x → z ∈ y

Proof

Step Hyp Ref Expression
1 equvinv ⊢ x = y ↔ ∃ t t = x ∧ t = y
2 ax9v2 ⊢ x = t → z ∈ x → z ∈ t
3 2 equcoms ⊢ t = x → z ∈ x → z ∈ t
4 ax9v1 ⊢ t = y → z ∈ t → z ∈ y
5 3 4 sylan9 ⊢ t = x ∧ t = y → z ∈ x → z ∈ y
6 5 exlimiv ⊢ ∃ t t = x ∧ t = y → z ∈ x → z ∈ y
7 1 6 sylbi ⊢ x = y → z ∈ x → z ∈ y