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 ) )