Theorem fnopabeqd

Description: Equality deduction for function abstractions. (Contributed by Jeff Madsen, 19-Jun-2011)

	Ref	Expression
Hypothesis	fnopabeqd.1	⊢ ( 𝜑 → 𝐵 = 𝐶 )
Assertion	fnopabeqd	⊢ ( 𝜑 → { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵 ) } = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶 ) } )

Proof

Step	Hyp	Ref	Expression
1		fnopabeqd.1	⊢ ( 𝜑 → 𝐵 = 𝐶 )
2	1	eqeq2d	⊢ ( 𝜑 → ( 𝑦 = 𝐵 ↔ 𝑦 = 𝐶 ) )
3	2	anbi2d	⊢ ( 𝜑 → ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶 ) ) )
4	3	opabbidv	⊢ ( 𝜑 → { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵 ) } = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶 ) } )