Theorem ceqsrexbv 3655

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 3188	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ ∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑)))
2		eleq1 2826	. . . . . . 7 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
3	2	adantr 480	. . . . . 6 ⊢ ((𝑥 = 𝐴 ∧ 𝜑) → (𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
4	3	pm5.32ri 575	. . . . 5 ⊢ ((𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
5	4	bicomi 224	. . . 4 ⊢ ((𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
6	5	baib 535	. . 3 ⊢ (𝑥 ∈ 𝐵 → ((𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝑥 = 𝐴 ∧ 𝜑)))
7	6	rexbiia 3089	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ ∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑))
8		ceqsrexv.1	. . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
9	8	ceqsrexv 3654	. . 3 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))
10	9	pm5.32i 574	. 2 ⊢ ((𝐴 ∈ 𝐵 ∧ ∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓))
11	1, 7, 10	3bitr3i 301	1 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1536   ∈ wcel 2105  ∃wrex 3067

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1791  ax-4 1805  ax-5 1907  ax-6 1964  ax-7 2004  ax-8 2107  ax-9 2115  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1539  df-ex 1776  df-sb 2062  df-clab 2712  df-cleq 2726  df-clel 2813  df-rex 3068

This theorem is referenced by:  ceqsralbv  3656  marypha2lem2  9473  psdmul  22187  txkgen  23675  eq0rabdioph  42763