Theorem ceqsrexv 3609

Description: Elimination of a restricted existential quantifier, using implicit substitution. (Contributed by NM, 30-Apr-2004.)

Hypothesis

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

Assertion

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

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

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem ceqsrexv

Step	Hyp	Ref	Expression
1		df-rex 3061	. . 3 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
2		an12 645	. . . 4 ⊢ ((𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
3	2	exbii 1849	. . 3 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
4	1, 3	bitr4i 278	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)))
5		eleq1 2824	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
6		ceqsrexv.1	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
7	5, 6	anbi12d 632	. . . 4 ⊢ (𝑥 = 𝐴 → ((𝑥 ∈ 𝐵 ∧ 𝜑) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓)))
8	7	ceqsexgv 3608	. . 3 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ (𝐴 ∈ 𝐵 ∧ 𝜓)))
9	8	bianabs 541	. 2 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ 𝜓))
10	4, 9	bitrid 283	1 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1541  ∃wex 1780   ∈ wcel 2113  ∃wrex 3060

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 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2715  df-cleq 2728  df-clel 2811  df-rex 3061

This theorem is referenced by:  ceqsrexbv  3610  ceqsrex2v  3612  reuxfrd  3706  f1oiso  7297  creur  12139  creui  12140  deg1ldg  26053  ulm2  26350  iscgra1  28882  reuxfrdf  32565  poimirlem24  37845  eqlkr3  39361  diclspsn  41454  rmxdiophlem  43257  expdiophlem1  43263  expdiophlem2  43264