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 = y y = x
4 2 3 sylg x x = y y y = x