Theorem reurab 3695

Description: Restricted existential uniqueness of a restricted abstraction. (Contributed by Scott Fenton, 8-Aug-2024.)

Hypothesis

Ref	Expression
reurab.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
reurab	⊢ (∃!𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓}𝜒 ↔ ∃!𝑥 ∈ 𝐴 (𝜑 ∧ 𝜒))

Distinct variable groups:   𝑦,𝐴   𝜑,𝑦   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥,𝑦)   𝜒(𝑥,𝑦)   𝐴(𝑥)

Proof of Theorem reurab

Step	Hyp	Ref	Expression
1		reurab.1	. . . . . . . 8 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	bicomd 222	. . . . . . 7 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜑))
3	2	equcoms 2016	. . . . . 6 ⊢ (𝑦 = 𝑥 → (𝜓 ↔ 𝜑))
4	3	elrab 3681	. . . . 5 ⊢ (𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓} ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
5	4	anbi1i 622	. . . 4 ⊢ ((𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓} ∧ 𝜒) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒))
6		anass 467	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
7	5, 6	bitri 274	. . 3 ⊢ ((𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓} ∧ 𝜒) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
8	7	eubii 2574	. 2 ⊢ (∃!𝑥(𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓} ∧ 𝜒) ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
9		df-reu 3365	. 2 ⊢ (∃!𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓}𝜒 ↔ ∃!𝑥(𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓} ∧ 𝜒))
10		df-reu 3365	. 2 ⊢ (∃!𝑥 ∈ 𝐴 (𝜑 ∧ 𝜒) ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
11	8, 9, 10	3bitr4i 302	1 ⊢ (∃!𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓}𝜒 ↔ ∃!𝑥 ∈ 𝐴 (𝜑 ∧ 𝜒))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 394   ∈ wcel 2099  ∃!weu 2557  ∃!wreu 3362  {crab 3419

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2697

This theorem depends on definitions:  df-bi 206  df-an 395  df-tru 1537  df-ex 1775  df-sb 2061  df-mo 2529  df-eu 2558  df-clab 2704  df-cleq 2718  df-clel 2803  df-reu 3365  df-rab 3420  df-v 3464

This theorem is referenced by:  eqscut  27835