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 ¬ z z = x ¬ z z = y x = y z x = y

Proof

Step Hyp Ref Expression
1 nfeqf ¬ z z = x ¬ z z = y z x = y
2 1 nf5rd ¬ z z = x ¬ z z = y x = y z x = y
3 2 ex ¬ z z = x ¬ z z = y x = y z x = y