Metamath Proof Explorer


Theorem ax13fromc9

Description: Derive ax-13 from ax-c9 and other older axioms.

This proof uses newer axioms ax-4 and ax-6 , but since these are proved from the older axioms above, this is acceptable and lets us avoid having to reprove several earlier theorems to use ax-c4 and ax-c10 . (Contributed by NM, 21-Dec-2015) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion ax13fromc9 ¬x=yy=zxy=z

Proof

Step Hyp Ref Expression
1 ax-c5 xx=yx=y
2 1 con3i ¬x=y¬xx=y
3 ax-c5 xx=zx=z
4 3 con3i ¬x=z¬xx=z
5 ax-c9 ¬xx=y¬xx=zy=zxy=z
6 2 4 5 syl2im ¬x=y¬x=zy=zxy=z
7 ax13b ¬x=yy=zxy=z¬x=y¬x=zy=zxy=z
8 6 7 mpbir ¬x=yy=zxy=z