Theorem fnopabeqd 35158

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 2809	. . 3 ⊢ (𝜑 → (𝑦 = 𝐵 ↔ 𝑦 = 𝐶))
3	2	anbi2d 631	. 2 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶)))
4	3	opabbidv 5096	1 ⊢ (𝜑 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶)})

Syntax hints:   → wi 4   ∧ wa 399   = wceq 1538   ∈ wcel 2111  {copab 5092

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-9 2121  ax-12 2175  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-nf 1786  df-sb 2070  df-clab 2777  df-cleq 2791  df-opab 5093

This theorem is referenced by: (None)