Theorem dfsb2ALT 2591

Description: Alternate version of dfsb2 2532. (Contributed by NM, 17-Feb-2005.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
dfsb1.ph	⊢ (𝜃 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))

Assertion

Ref	Expression
dfsb2ALT	⊢ (𝜃 ↔ ((𝑥 = 𝑦 ∧ 𝜑) ∨ ∀𝑥(𝑥 = 𝑦 → 𝜑)))

Proof of Theorem dfsb2ALT

Step	Hyp	Ref	Expression
1		sp 2182	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑦 → 𝑥 = 𝑦)
2		dfsb1.ph	. . . . . 6 ⊢ (𝜃 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
3	2	sbequ2ALT 2580	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜃 → 𝜑))
4	3	sps 2184	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑦 → (𝜃 → 𝜑))
5		orc 863	. . . 4 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → ((𝑥 = 𝑦 ∧ 𝜑) ∨ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
6	1, 4, 5	syl6an 682	. . 3 ⊢ (∀𝑥 𝑥 = 𝑦 → (𝜃 → ((𝑥 = 𝑦 ∧ 𝜑) ∨ ∀𝑥(𝑥 = 𝑦 → 𝜑))))
7	2	sb4ALT 2588	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝜃 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
8		olc 864	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → ((𝑥 = 𝑦 ∧ 𝜑) ∨ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
9	7, 8	syl6 35	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝜃 → ((𝑥 = 𝑦 ∧ 𝜑) ∨ ∀𝑥(𝑥 = 𝑦 → 𝜑))))
10	6, 9	pm2.61i 184	. 2 ⊢ (𝜃 → ((𝑥 = 𝑦 ∧ 𝜑) ∨ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
11	2	sbequ1ALT 2579	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 → 𝜃))
12	11	imp 409	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝜃)
13	2	sb2ALT 2587	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → 𝜃)
14	12, 13	jaoi 853	. 2 ⊢ (((𝑥 = 𝑦 ∧ 𝜑) ∨ ∀𝑥(𝑥 = 𝑦 → 𝜑)) → 𝜃)
15	10, 14	impbii 211	1 ⊢ (𝜃 ↔ ((𝑥 = 𝑦 ∧ 𝜑) ∨ ∀𝑥(𝑥 = 𝑦 → 𝜑)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 398   ∨ wo 843  ∀wal 1535  ∃wex 1780

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-10 2145  ax-12 2177  ax-13 2390

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1781  df-nf 1785

This theorem is referenced by:  dfsb3ALT  2592