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. Usage of this theorem has been discouraged later on to avoid ax-13 propagation. Check out bj-axc10v for a weaker version requiring less axioms. (Contributed by NM, 5-Aug-1993) (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Assertion	axc10	$$⊢\forall x\left(x=y\to \forall x\phi \right)\to \phi$$

Proof

Step	Hyp	Ref	Expression
1		ax6	$$⊢\neg \forall x\neg x=y$$
2		con3	$$⊢\left(x=y\to \forall x\phi \right)\to \left(\neg \forall x\phi \to \neg x=y\right)$$
3	2	al2imi	$$⊢\forall x\left(x=y\to \forall x\phi \right)\to \left(\forall x\neg \forall x\phi \to \forall x\neg x=y\right)$$
4	1 3	mtoi	$$⊢\forall x\left(x=y\to \forall x\phi \right)\to \neg \forall x\neg \forall x\phi$$
5		axc7	$$⊢\neg \forall x\neg \forall x\phi \to \phi$$
6	4 5	syl	$$⊢\forall x\left(x=y\to \forall x\phi \right)\to \phi$$