Metamath Proof Explorer


Theorem ax6fromc10

Description: Rederivation of Axiom ax-6 from ax-c7 , ax-c10 , ax-gen and propositional calculus. See axc10 for the derivation of ax-c10 from ax-6 . Lemma L18 in Megill p. 446 (p. 14 of the preprint). (Contributed by NM, 14-May-1993) (Proof modification is discouraged.) Use ax-6 instead. (New usage is discouraged.)

Ref Expression
Assertion ax6fromc10
|- -. A. x -. x = y

Proof

Step Hyp Ref Expression
1 ax-c10
 |-  ( A. x ( x = y -> A. x -. A. x -. x = y ) -> -. A. x -. x = y )
2 ax-c7
 |-  ( -. A. x -. A. x -. x = y -> -. x = y )
3 2 con4i
 |-  ( x = y -> A. x -. A. x -. x = y )
4 1 3 mpg
 |-  -. A. x -. x = y