Metamath Proof Explorer


Theorem axc11n11

Description: Proof of axc11n from { ax-1 -- ax-7 , axc11 } . Almost identical to axc11nfromc11 . (Contributed by NM, 6-Jul-2021) (Proof modification is discouraged.)

Ref Expression
Assertion axc11n11 ( ∀ 𝑥 𝑥 = 𝑦 → ∀ 𝑦 𝑦 = 𝑥 )

Proof

Step Hyp Ref Expression
1 axc11 ( ∀ 𝑥 𝑥 = 𝑦 → ( ∀ 𝑥 𝑥 = 𝑦 → ∀ 𝑦 𝑥 = 𝑦 ) )
2 1 pm2.43i ( ∀ 𝑥 𝑥 = 𝑦 → ∀ 𝑦 𝑥 = 𝑦 )
3 equcomi ( 𝑥 = 𝑦𝑦 = 𝑥 )
4 2 3 sylg ( ∀ 𝑥 𝑥 = 𝑦 → ∀ 𝑦 𝑦 = 𝑥 )