Theorem ax12indi 38945

Description: Induction step for constructing a substitution instance of ax-c15 38890 without using ax-c15 38890. Implication case. (Contributed by NM, 21-Jan-2007.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypotheses

Ref	Expression
ax12indn.1	⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
ax12indi.2	⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (𝜓 → ∀𝑥(𝑥 = 𝑦 → 𝜓))))

Assertion

Ref	Expression
ax12indi	⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → ((𝜑 → 𝜓) → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)))))

Proof of Theorem ax12indi

Step	Hyp	Ref	Expression
1		ax12indn.1	. . . . . 6 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
2	1	ax12indn 38944	. . . . 5 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (¬ 𝜑 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑))))
3	2	imp 406	. . . 4 ⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∧ 𝑥 = 𝑦) → (¬ 𝜑 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
4		pm2.21 123	. . . . . 6 ⊢ (¬ 𝜑 → (𝜑 → 𝜓))
5	4	imim2i 16	. . . . 5 ⊢ ((𝑥 = 𝑦 → ¬ 𝜑) → (𝑥 = 𝑦 → (𝜑 → 𝜓)))
6	5	alimi 1811	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → ¬ 𝜑) → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)))
7	3, 6	syl6 35	. . 3 ⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∧ 𝑥 = 𝑦) → (¬ 𝜑 → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓))))
8		ax12indi.2	. . . . 5 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (𝜓 → ∀𝑥(𝑥 = 𝑦 → 𝜓))))
9	8	imp 406	. . . 4 ⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∧ 𝑥 = 𝑦) → (𝜓 → ∀𝑥(𝑥 = 𝑦 → 𝜓)))
10		ax-1 6	. . . . . 6 ⊢ (𝜓 → (𝜑 → 𝜓))
11	10	imim2i 16	. . . . 5 ⊢ ((𝑥 = 𝑦 → 𝜓) → (𝑥 = 𝑦 → (𝜑 → 𝜓)))
12	11	alimi 1811	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)))
13	9, 12	syl6 35	. . 3 ⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∧ 𝑥 = 𝑦) → (𝜓 → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓))))
14	7, 13	jad 187	. 2 ⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∧ 𝑥 = 𝑦) → ((𝜑 → 𝜓) → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓))))
15	14	ex 412	1 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → ((𝜑 → 𝜓) → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)))))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395  ∀wal 1538

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-10 2141  ax-12 2177

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780

This theorem is referenced by: (None)