Theorem pairreueq 48418

Description: Two equivalent representations of the existence of a unique proper pair. (Contributed by AV, 1-Mar-2023.)

Hypothesis

Ref	Expression
pairreueq.p	⊢ 𝑃 = {𝑥 ∈ 𝒫 𝑉 ∣ (♯‘𝑥) = 2}

Assertion

Ref	Expression
pairreueq	⊢ (∃!𝑝 ∈ 𝑃 𝜑 ↔ ∃!𝑝 ∈ 𝒫 𝑉((♯‘𝑝) = 2 ∧ 𝜑))

Distinct variable groups:   𝑥,𝑝   𝑥,𝑉

Allowed substitution hints:   𝜑(𝑥, 𝑝)   𝑃(𝑥, 𝑝)   𝑉(𝑝)

Proof of Theorem pairreueq

Step	Hyp	Ref	Expression
1		fveqeq2 6891	. . . . . 6 ⊢ (𝑥 = 𝑝 → ((♯‘𝑥) = 2 ↔ (♯‘𝑝) = 2))
2		pairreueq.p	. . . . . 6 ⊢ 𝑃 = {𝑥 ∈ 𝒫 𝑉 ∣ (♯‘𝑥) = 2}
3	1, 2	elrab2 3652	. . . . 5 ⊢ (𝑝 ∈ 𝑃 ↔ (𝑝 ∈ 𝒫 𝑉 ∧ (♯‘𝑝) = 2))
4	3	anbi1i 636	. . . 4 ⊢ ((𝑝 ∈ 𝑃 ∧ 𝜑) ↔ ((𝑝 ∈ 𝒫 𝑉 ∧ (♯‘𝑝) = 2) ∧ 𝜑))
5		anass 474	. . . 4 ⊢ (((𝑝 ∈ 𝒫 𝑉 ∧ (♯‘𝑝) = 2) ∧ 𝜑) ↔ (𝑝 ∈ 𝒫 𝑉 ∧ ((♯‘𝑝) = 2 ∧ 𝜑)))
6	4, 5	bitri 278	. . 3 ⊢ ((𝑝 ∈ 𝑃 ∧ 𝜑) ↔ (𝑝 ∈ 𝒫 𝑉 ∧ ((♯‘𝑝) = 2 ∧ 𝜑)))
7	6	eubii 2612	. 2 ⊢ (∃!𝑝(𝑝 ∈ 𝑃 ∧ 𝜑) ↔ ∃!𝑝(𝑝 ∈ 𝒫 𝑉 ∧ ((♯‘𝑝) = 2 ∧ 𝜑)))
8		df-reu 3368	. 2 ⊢ (∃!𝑝 ∈ 𝑃 𝜑 ↔ ∃!𝑝(𝑝 ∈ 𝑃 ∧ 𝜑))
9		df-reu 3368	. 2 ⊢ (∃!𝑝 ∈ 𝒫 𝑉((♯‘𝑝) = 2 ∧ 𝜑) ↔ ∃!𝑝(𝑝 ∈ 𝒫 𝑉 ∧ ((♯‘𝑝) = 2 ∧ 𝜑)))
10	7, 8, 9	3bitr4i 306	1 ⊢ (∃!𝑝 ∈ 𝑃 𝜑 ↔ ∃!𝑝 ∈ 𝒫 𝑉((♯‘𝑝) = 2 ∧ 𝜑))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  ∃!weu 2595  ∃!wreu 3365  {crab 3414  𝒫 cpw 4560  ‘cfv 6537  2c2 12323  ♯chash 14398

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 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-reu 3368  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-br 5108  df-iota 6493  df-fv 6545

This theorem is used by:  requad2  48547