Theorem ax12indn 39758

Description: Induction step for constructing a substitution instance of ax-c15 39704 without using ax-c15 39704. 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 2220	. . 3 ⊢ ((𝑥 = 𝑦 ∧ ¬ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ ¬ 𝜑))
2		exanali 1892	. . . 4 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ ¬ 𝜑) ↔ ¬ ∀𝑥(𝑥 = 𝑦 → 𝜑))
3		hbn1 2180	. . . . 5 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥 ¬ ∀𝑥 𝑥 = 𝑦)
4		hbn1 2180	. . . . 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 154	. . . . . . 7 ⊢ ((𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) → (¬ ∀𝑥(𝑥 = 𝑦 → 𝜑) → ¬ 𝜑))
7	5, 6	syl6 36	. . . . . 6 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (¬ ∀𝑥(𝑥 = 𝑦 → 𝜑) → ¬ 𝜑)))
8	7	com23 87	. . . . 5 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (¬ ∀𝑥(𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → ¬ 𝜑)))
9	3, 4, 8	alrimdh 1896	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (¬ ∀𝑥(𝑥 = 𝑦 → 𝜑) → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
10	2, 9	biimtrid 245	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦 ∧ ¬ 𝜑) → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
11	1, 10	syl5 35	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑦 ∧ ¬ 𝜑) → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
12	11	expd 421	1 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (¬ 𝜑 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑))))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401  ∀wal 1568  ∃wex 1812

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2179  ax-12 2216

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813

This theorem is used by:  ax12indi  39759