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 = y -> ( y = z -> A. x y = z ) )

Proof

Step Hyp Ref Expression
1 ax-c5
 |-  ( A. x x = y -> x = y )
2 1 con3i
 |-  ( -. x = y -> -. A. x x = y )
3 ax-c5
 |-  ( A. x x = z -> x = z )
4 3 con3i
 |-  ( -. x = z -> -. A. x x = z )
5 ax-c9
 |-  ( -. A. x x = y -> ( -. A. x x = z -> ( y = z -> A. x y = z ) ) )
6 2 4 5 syl2im
 |-  ( -. x = y -> ( -. x = z -> ( y = z -> A. x y = z ) ) )
7 ax13b
 |-  ( ( -. x = y -> ( y = z -> A. x y = z ) ) <-> ( -. x = y -> ( -. x = z -> ( y = z -> A. x y = z ) ) ) )
8 6 7 mpbir
 |-  ( -. x = y -> ( y = z -> A. x y = z ) )