Theorem ceqsrexbv 3610

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 3194	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ ∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑)))
2		eleq1 2848	. . . . . . 7 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
3	2	adantr 486	. . . . . 6 ⊢ ((𝑥 = 𝐴 ∧ 𝜑) → (𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
4	3	pm5.32ri 586	. . . . 5 ⊢ ((𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
5	4	bicomi 227	. . . 4 ⊢ ((𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
6	5	baib 545	. . 3 ⊢ (𝑥 ∈ 𝐵 → ((𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝑥 = 𝐴 ∧ 𝜑)))
7	6	rexbiia 3107	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝐴 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ ∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑))
8		ceqsrexv.1	. . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
9	8	ceqsrexv 3609	. . 3 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))
10	9	pm5.32i 585	. 2 ⊢ ((𝐴 ∈ 𝐵 ∧ ∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓))
11	1, 7, 10	3bitr3i 304	1 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  ∃wrex 3086

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-rex 3087

This theorem is used by:  ceqsralbv  3611  marypha2lem2  9407  psdmul  22395  txkgen  23879  eq0rabdioph  43622