Theorem pairreueq

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

	Ref	Expression
Hypothesis	pairreueq.p	⊢ 𝑃 = { 𝑥 ∈ 𝒫 𝑉 ∣ ( ♯ ‘ 𝑥 ) = 2 }
Assertion	pairreueq	⊢ ( ∃! 𝑝 ∈ 𝑃 𝜑 ↔ ∃! 𝑝 ∈ 𝒫 𝑉 ( ( ♯ ‘ 𝑝 ) = 2 ∧ 𝜑 ) )

Proof

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