Metamath Proof Explorer


Theorem axc9

Description: Derive set.mm's original ax-c9 from the shorter ax-13 . Usage is discouraged to avoid uninformed ax-13 propagation. (Contributed by NM, 29-Nov-2015) (Revised by NM, 24-Dec-2015) (Proof shortened by Wolf Lammen, 29-Apr-2018) (New usage is discouraged.)

Ref Expression
Assertion axc9
|- ( -. A. z z = x -> ( -. A. z z = y -> ( x = y -> A. z x = y ) ) )

Proof

Step Hyp Ref Expression
1 nfeqf
 |-  ( ( -. A. z z = x /\ -. A. z z = y ) -> F/ z x = y )
2 1 nf5rd
 |-  ( ( -. A. z z = x /\ -. A. z z = y ) -> ( x = y -> A. z x = y ) )
3 2 ex
 |-  ( -. A. z z = x -> ( -. A. z z = y -> ( x = y -> A. z x = y ) ) )