Theorem reurab 3659

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 226	. . . . . . 7 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜑))
3	2	equcoms 2053	. . . . . 6 ⊢ (𝑦 = 𝑥 → (𝜓 ↔ 𝜑))
4	3	elrab 3645	. . . . 5 ⊢ (𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓} ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
5	4	anbi1i 636	. . . 4 ⊢ ((𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓} ∧ 𝜒) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒))
6		anass 474	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ 𝜒) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
7	5, 6	bitri 278	. . 3 ⊢ ((𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓} ∧ 𝜒) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
8	7	eubii 2611	. 2 ⊢ (∃!𝑥(𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓} ∧ 𝜒) ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
9		df-reu 3367	. 2 ⊢ (∃!𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓}𝜒 ↔ ∃!𝑥(𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓} ∧ 𝜒))
10		df-reu 3367	. 2 ⊢ (∃!𝑥 ∈ 𝐴 (𝜑 ∧ 𝜒) ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜒)))
11	8, 9, 10	3bitr4i 306	1 ⊢ (∃!𝑥 ∈ {𝑦 ∈ 𝐴 ∣ 𝜓}𝜒 ↔ ∃!𝑥 ∈ 𝐴 (𝜑 ∧ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∈ wcel 2145  ∃!weu 2594  ∃!wreu 3364  {crab 3413

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-mo 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-reu 3367  df-rab 3414  df-v 3453

This theorem is used by:  eqcuts  28164