Metamath Proof Explorer


Theorem axc10

Description: Show that the original axiom ax-c10 can be derived from ax6 and axc7 (on top of propositional calculus, ax-gen , and ax-4 ). See ax6fromc10 for the rederivation of ax6 from ax-c10 .

Normally, axc10 should be used rather than ax-c10 , except by theorems specifically studying the latter's properties. See bj-axc10v for a weaker version requiring fewer axioms. (Contributed by NM, 5-Aug-1993) (Proof modification is discouraged.) Usage of this theorem is discouraged because it depends on ax-13 . (New usage is discouraged.)

Ref Expression
Assertion axc10 x x = y x φ φ

Proof

Step Hyp Ref Expression
1 ax6 ¬ x ¬ x = y
2 con3 x = y x φ ¬ x φ ¬ x = y
3 2 al2imi x x = y x φ x ¬ x φ x ¬ x = y
4 1 3 mtoi x x = y x φ ¬ x ¬ x φ
5 axc7 ¬ x ¬ x φ φ
6 4 5 syl x x = y x φ φ