Theorem sbi1ALT 2605

Description: Alternate version of sbi1 2075. (Contributed by NM, 14-May-1993.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypotheses

Ref	Expression
dfsb1.p5	⊢ (𝜃 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
dfsb1.s2	⊢ (𝜏 ↔ ((𝑥 = 𝑦 → 𝜓) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
dfsb1.im	⊢ (𝜂 ↔ ((𝑥 = 𝑦 → (𝜑 → 𝜓)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝜑 → 𝜓))))

Assertion

Ref	Expression
sbi1ALT	⊢ (𝜂 → (𝜃 → 𝜏))

Proof of Theorem sbi1ALT

Step	Hyp	Ref	Expression
1		dfsb1.p5	. . . . . 6 ⊢ (𝜃 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
2	1	sbequ2ALT 2579	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜃 → 𝜑))
3		dfsb1.im	. . . . . 6 ⊢ (𝜂 ↔ ((𝑥 = 𝑦 → (𝜑 → 𝜓)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝜑 → 𝜓))))
4	3	sbequ2ALT 2579	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜂 → (𝜑 → 𝜓)))
5	2, 4	syl5d 73	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜂 → (𝜃 → 𝜓)))
6		dfsb1.s2	. . . . 5 ⊢ (𝜏 ↔ ((𝑥 = 𝑦 → 𝜓) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
7	6	sbequ1ALT 2578	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜓 → 𝜏))
8	5, 7	syl6d 75	. . 3 ⊢ (𝑥 = 𝑦 → (𝜂 → (𝜃 → 𝜏)))
9	8	sps 2183	. 2 ⊢ (∀𝑥 𝑥 = 𝑦 → (𝜂 → (𝜃 → 𝜏)))
10	1	sb4ALT 2587	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝜃 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
11	3	sb4ALT 2587	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝜂 → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓))))
12		ax-2 7	. . . . . 6 ⊢ ((𝑥 = 𝑦 → (𝜑 → 𝜓)) → ((𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → 𝜓)))
13	12	al2imi 1815	. . . . 5 ⊢ (∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)) → (∀𝑥(𝑥 = 𝑦 → 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜓)))
14	6	sb2ALT 2586	. . . . 5 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) → 𝜏)
15	13, 14	syl6 35	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)) → (∀𝑥(𝑥 = 𝑦 → 𝜑) → 𝜏))
16	11, 15	syl6 35	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝜂 → (∀𝑥(𝑥 = 𝑦 → 𝜑) → 𝜏)))
17	10, 16	syl5d 73	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝜂 → (𝜃 → 𝜏)))
18	9, 17	pm2.61i 184	1 ⊢ (𝜂 → (𝜃 → 𝜏))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 398  ∀wal 1534  ∃wex 1779

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 1969  ax-7 2014  ax-10 2144  ax-12 2176  ax-13 2389

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1780  df-nf 1784

This theorem is referenced by:  sbimALT  2607