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 3088	. . 3 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
2		an12 658	. . . 4 ⊢ ((𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
3	2	exbii 1881	. . 3 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)))
4	1, 3	bitr4i 281	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)))
5		eleq1 2849	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
6		ceqsrexv.1	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
7	5, 6	anbi12d 644	. . . 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 551	. 2 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥(𝑥 = 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ 𝜓))
10	4, 9	bitrid 286	1 ⊢ (𝐴 ∈ 𝐵 → (∃𝑥 ∈ 𝐵 (𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570  ∃wex 1812   ∈ wcel 2145  ∃wrex 3087

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 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-rex 3088

This theorem is used by:  ceqsrexbv  3610  ceqsrex2v  3612  reuxfrd  3706  f1oiso  7357  creur  12307  creui  12308  deg1ldg  26403  ulm2  26705  iscgra1  29310  reuxfrdf  33080  poimirlem24  38542  eqlkr3  40138  diclspsn  42231  rmxdiophlem  44001  expdiophlem1  44007  expdiophlem2  44008