Metamath Proof Explorer


Theorem ax9v2

Description: Second of two weakened versions of ax9v , with an extra disjoint variable condition on y , z see comments there. (Contributed by BJ, 7-Dec-2020)

Ref Expression
Assertion ax9v2 ( 𝑥 = 𝑦 → ( 𝑧 ∈ 𝑥 → 𝑧 ∈ 𝑦 ) )

Proof

Step Hyp Ref Expression
1 ax9v ⊢ ( 𝑥 = 𝑦 → ( 𝑧 ∈ 𝑥 → 𝑧 ∈ 𝑦 ) )