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 xx=yyy=x

Proof

Step Hyp Ref Expression
1 axc11 xx=yxx=yyx=y
2 1 pm2.43i xx=yyx=y
3 equcomi x=yy=x
4 2 3 sylg xx=yyy=x