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 3087	. . 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 2848	. . . . 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 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:  ceqsrexbv  3610  ceqsrex2v  3612  reuxfrd  3706  f1oiso  7352  creur  12236  creui  12237  deg1ldg  26317  ulm2  26621  iscgra1  29196  reuxfrdf  32966  poimirlem24  38393  eqlkr3  39974  diclspsn  42067  rmxdiophlem  43856  expdiophlem1  43862  expdiophlem2  43863