Theorem ax12indn 36884

Description: Induction step for constructing a substitution instance of ax-c15 36830 without using ax-c15 36830. Negation case. (Contributed by NM, 21-Jan-2007.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
ax12indn.1	⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))

Assertion

Ref	Expression
ax12indn	⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (¬ 𝜑 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑))))

Proof of Theorem ax12indn

Step	Hyp	Ref	Expression
1		19.8a 2176	. . 3 ⊢ ((𝑥 = 𝑦 ∧ ¬ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ ¬ 𝜑))
2		exanali 1863	. . . 4 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ ¬ 𝜑) ↔ ¬ ∀𝑥(𝑥 = 𝑦 → 𝜑))
3		hbn1 2140	. . . . 5 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥 ¬ ∀𝑥 𝑥 = 𝑦)
4		hbn1 2140	. . . . 5 ⊢ (¬ ∀𝑥(𝑥 = 𝑦 → 𝜑) → ∀𝑥 ¬ ∀𝑥(𝑥 = 𝑦 → 𝜑))
5		ax12indn.1	. . . . . . 7 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
6		con3 153	. . . . . . 7 ⊢ ((𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) → (¬ ∀𝑥(𝑥 = 𝑦 → 𝜑) → ¬ 𝜑))
7	5, 6	syl6 35	. . . . . 6 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (¬ ∀𝑥(𝑥 = 𝑦 → 𝜑) → ¬ 𝜑)))
8	7	com23 86	. . . . 5 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (¬ ∀𝑥(𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → ¬ 𝜑)))
9	3, 4, 8	alrimdh 1867	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (¬ ∀𝑥(𝑥 = 𝑦 → 𝜑) → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
10	2, 9	syl5bi 241	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦 ∧ ¬ 𝜑) → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
11	1, 10	syl5 34	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑦 ∧ ¬ 𝜑) → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
12	11	expd 415	1 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (¬ 𝜑 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑))))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395  ∀wal 1537  ∃wex 1783

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-10 2139  ax-12 2173

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1784

This theorem is referenced by:  ax12indi  36885