Metamath Proof Explorer


Theorem axc16nfALT

Description: Alternate proof of axc16nf , shorter but requiring ax-11 and ax-13 . (Contributed by Mario Carneiro, 7-Oct-2016) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion axc16nfALT ( ∀ 𝑥 𝑥 = 𝑦 → Ⅎ 𝑧 𝜑 )

Proof

Step Hyp Ref Expression
1 nfae 𝑧𝑥 𝑥 = 𝑦
2 axc16g ( ∀ 𝑥 𝑥 = 𝑦 → ( 𝜑 → ∀ 𝑧 𝜑 ) )
3 1 2 nf5d ( ∀ 𝑥 𝑥 = 𝑦 → Ⅎ 𝑧 𝜑 )