Theorem ceqsrexbv 2880

Description: Elimination of a restricted existential quantifier, using implicit substitution. (Contributed by Mario Carneiro, 14-Mar-2014.)

Hypothesis

Ref	Expression
ceqsrexv.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
ceqsrexbv	⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜓,𝑥

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem ceqsrexbv

Step	Hyp	Ref	Expression
1		r19.42v 2644	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ ∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑)))
2		eleq1 2250	. . . . . . 7 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
3	2	adantr 276	. . . . . 6 ⊢ ((𝑥 = 𝐴 ∧ 𝜑) → (𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
4	3	pm5.32ri 455	. . . . 5 ⊢ ((𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
5	4	bicomi 132	. . . 4 ⊢ ((𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
6	5	baib 920	. . 3 ⊢ (𝑥 ∈ 𝐵 → ((𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝑥 = 𝐴 ∧ 𝜑)))
7	6	rexbiia 2502	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ ∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑))
8		ceqsrexv.1	. . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
9	8	ceqsrexv 2879	. . 3 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))
10	9	pm5.32i 454	. 2 ⊢ ((𝐴 ∈ 𝐵 ∧ ∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓))
11	1, 7, 10	3bitr3i 210	1 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1363   ∈ wcel 2158  ∃wrex 2466

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1457  ax-7 1458  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-10 1515  ax-11 1516  ax-i12 1517  ax-bndl 1519  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544  ax-i5r 1545  ax-ext 2169

This theorem depends on definitions:  df-bi 117  df-tru 1366  df-nf 1471  df-sb 1773  df-clab 2174  df-cleq 2180  df-clel 2183  df-nfc 2318  df-rex 2471  df-v 2751

This theorem is referenced by:  frecsuclem  6421