Theorem sbiedALT 2614

Description: Alternate version of sbied 2545. (Contributed by NM, 30-Jun-1994.) (Revised by Mario Carneiro, 4-Oct-2016.) (Proof shortened by Wolf Lammen, 24-Jun-2018.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypotheses

Ref	Expression
dfsb1.s5	⊢ (𝜏 ↔ ((𝑥 = 𝑦 → 𝜓) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
sbiedALT.1	⊢ Ⅎ𝑥𝜑
sbiedALT.2	⊢ (𝜑 → Ⅎ𝑥𝜒)
sbiedALT.3	⊢ (𝜑 → (𝑥 = 𝑦 → (𝜓 ↔ 𝜒)))

Assertion

Ref	Expression
sbiedALT	⊢ (𝜑 → (𝜏 ↔ 𝜒))

Proof of Theorem sbiedALT

Step	Hyp	Ref	Expression
1		dfsb1.s5	. . . 4 ⊢ (𝜏 ↔ ((𝑥 = 𝑦 → 𝜓) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
2		biid 263	. . . 4 ⊢ (((𝑥 = 𝑦 → (𝜑 → 𝜓)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝜑 → 𝜓))) ↔ ((𝑥 = 𝑦 → (𝜑 → 𝜓)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝜑 → 𝜓))))
3		sbiedALT.1	. . . 4 ⊢ Ⅎ𝑥𝜑
4	1, 2, 3	sbrimALT 2609	. . 3 ⊢ (((𝑥 = 𝑦 → (𝜑 → 𝜓)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝜑 → 𝜓))) ↔ (𝜑 → 𝜏))
5		sbiedALT.2	. . . . 5 ⊢ (𝜑 → Ⅎ𝑥𝜒)
6	3, 5	nfim1 2199	. . . 4 ⊢ Ⅎ𝑥(𝜑 → 𝜒)
7		sbiedALT.3	. . . . . 6 ⊢ (𝜑 → (𝑥 = 𝑦 → (𝜓 ↔ 𝜒)))
8	7	com12 32	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 → (𝜓 ↔ 𝜒)))
9	8	pm5.74d 275	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝜑 → 𝜓) ↔ (𝜑 → 𝜒)))
10	2, 6, 9	sbieALT 2613	. . 3 ⊢ (((𝑥 = 𝑦 → (𝜑 → 𝜓)) ∧ ∃𝑥(𝑥 = 𝑦 ∧ (𝜑 → 𝜓))) ↔ (𝜑 → 𝜒))
11	4, 10	bitr3i 279	. 2 ⊢ ((𝜑 → 𝜏) ↔ (𝜑 → 𝜒))
12	11	pm5.74ri 274	1 ⊢ (𝜑 → (𝜏 ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398  ∃wex 1780  Ⅎwnf 1784

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:  sbco2ALT  2615