Theorem fnopabeqd 35922

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

Hypothesis

Ref	Expression
fnopabeqd.1	⊢ (𝜑 → 𝐵 = 𝐶)

Assertion

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

Distinct variable groups:   𝜑,𝑥   𝜑,𝑦

Allowed substitution hints:   𝐴(𝑥,𝑦)   𝐵(𝑥,𝑦)   𝐶(𝑥,𝑦)

Proof of Theorem fnopabeqd

Step	Hyp	Ref	Expression
1		fnopabeqd.1	. . . 4 ⊢ (𝜑 → 𝐵 = 𝐶)
2	1	eqeq2d 2747	. . 3 ⊢ (𝜑 → (𝑦 = 𝐵 ↔ 𝑦 = 𝐶))
3	2	anbi2d 630	. 2 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶)))
4	3	opabbidv 5147	1 ⊢ (𝜑 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶)})

Syntax hints:   → wi 4   ∧ wa 397   = wceq 1539   ∈ wcel 2104  {copab 5143

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-9 2114  ax-ext 2707

This theorem depends on definitions:  df-bi 206  df-an 398  df-ex 1780  df-sb 2066  df-clab 2714  df-cleq 2728  df-opab 5144

This theorem is referenced by: (None)