Metamath Proof Explorer


Theorem equcomi1

Description: Proof of equcomi from equid1 , avoiding use of ax-5 (the only use of ax-5 is via ax7 , so using ax-7 instead would remove dependency on ax-5 ). (Contributed by BJ, 8-Jul-2021) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion equcomi1
|- ( x = y -> y = x )

Proof

Step Hyp Ref Expression
1 equid1
 |-  x = x
2 ax7
 |-  ( x = y -> ( x = x -> y = x ) )
3 1 2 mpi
 |-  ( x = y -> y = x )