Metamath Proof Explorer


Theorem ax8v2

Description: Second of two weakened versions of ax8v , with an extra disjoint variable condition on y , z see comments there. (Contributed by BJ, 7-Dec-2020)

Ref Expression
Assertion ax8v2 ( 𝑥 = 𝑦 → ( 𝑥 ∈ 𝑧 → 𝑦 ∈ 𝑧 ) )

Proof

Step Hyp Ref Expression
1 ax8v ⊢ ( 𝑥 = 𝑦 → ( 𝑥 ∈ 𝑧 → 𝑦 ∈ 𝑧 ) )