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Metamath Proof Explorer

Theorem ax12indn

Description: Induction step for constructing a substitution instance of ax-c15 without using ax-c15 . Negation case. (Contributed by NM, 21-Jan-2007) (Proof modification is discouraged.) (New usage is discouraged.)

		Ref	Expression
	Hypothesis	ax12indn.1	$⊢ \neg \forall x x = y \to (x = y \to (φ \to \forall x (x = y \to φ)))$
	Assertion	ax12indn	$⊢ \neg \forall x x = y \to (x = y \to (\neg φ \to \forall x (x = y \to \neg φ)))$

Proof

Step	Hyp	Ref	Expression
1		ax12indn.1	$⊢ \neg \forall x x = y \to (x = y \to (φ \to \forall x (x = y \to φ)))$
2		19.8a	$⊢ (x = y \land \neg φ) \to \exists x (x = y \land \neg φ)$
3		exanali	$⊢ \exists x (x = y \land \neg φ) \leftrightarrow \neg \forall x (x = y \to φ)$
4		hbn1	$⊢ \neg \forall x x = y \to \forall x \neg \forall x x = y$
5		hbn1	$⊢ \neg \forall x (x = y \to φ) \to \forall x \neg \forall x (x = y \to φ)$
6		con3	$⊢ (φ \to \forall x (x = y \to φ)) \to (\neg \forall x (x = y \to φ) \to \neg φ)$
7	1 6	syl6	$⊢ \neg \forall x x = y \to (x = y \to (\neg \forall x (x = y \to φ) \to \neg φ))$
8	7	com23	$⊢ \neg \forall x x = y \to (\neg \forall x (x = y \to φ) \to (x = y \to \neg φ))$
9	4 5 8	alrimdh	$⊢ \neg \forall x x = y \to (\neg \forall x (x = y \to φ) \to \forall x (x = y \to \neg φ))$
10	3 9	biimtrid	$⊢ \neg \forall x x = y \to (\exists x (x = y \land \neg φ) \to \forall x (x = y \to \neg φ))$
11	2 10	syl5	$⊢ \neg \forall x x = y \to ((x = y \land \neg φ) \to \forall x (x = y \to \neg φ))$
12	11	expd	$⊢ \neg \forall x x = y \to (x = y \to (\neg φ \to \forall x (x = y \to \neg φ)))$