Theorem cbvopab2davw 36208

Description: Change the second bound variable in an ordered-pair class abstraction. Deduction form. (Contributed by GG, 14-Aug-2025.)

Hypothesis

Ref	Expression
cbvopab2davw.1	⊢ ((𝜑 ∧ 𝑦 = 𝑧) → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
cbvopab2davw	⊢ (𝜑 → {⟨𝑥, 𝑦⟩ ∣ 𝜓} = {⟨𝑥, 𝑧⟩ ∣ 𝜒})

Distinct variable groups:   𝜑,𝑥,𝑦,𝑧   𝜓,𝑧   𝜒,𝑦

Allowed substitution hints:   𝜓(𝑥,𝑦)   𝜒(𝑥,𝑧)

Proof of Theorem cbvopab2davw

Dummy variable 𝑡 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		opeq2 4881	. . . . . . . 8 ⊢ (𝑦 = 𝑧 → ⟨𝑥, 𝑦⟩ = ⟨𝑥, 𝑧⟩)
2	1	eqeq2d 2744	. . . . . . 7 ⊢ (𝑦 = 𝑧 → (𝑡 = ⟨𝑥, 𝑦⟩ ↔ 𝑡 = ⟨𝑥, 𝑧⟩))
3	2	adantl 481	. . . . . 6 ⊢ ((𝜑 ∧ 𝑦 = 𝑧) → (𝑡 = ⟨𝑥, 𝑦⟩ ↔ 𝑡 = ⟨𝑥, 𝑧⟩))
4		cbvopab2davw.1	. . . . . 6 ⊢ ((𝜑 ∧ 𝑦 = 𝑧) → (𝜓 ↔ 𝜒))
5	3, 4	anbi12d 631	. . . . 5 ⊢ ((𝜑 ∧ 𝑦 = 𝑧) → ((𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ (𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)))
6	5	cbvexdvaw 2034	. . . 4 ⊢ (𝜑 → (∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ ∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)))
7	6	exbidv 1917	. . 3 ⊢ (𝜑 → (∃𝑥∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ ∃𝑥∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)))
8	7	abbidv 2804	. 2 ⊢ (𝜑 → {𝑡 ∣ ∃𝑥∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)} = {𝑡 ∣ ∃𝑥∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)})
9		df-opab 5212	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜓} = {𝑡 ∣ ∃𝑥∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)}
10		df-opab 5212	. 2 ⊢ {⟨𝑥, 𝑧⟩ ∣ 𝜒} = {𝑡 ∣ ∃𝑥∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)}
11	8, 9, 10	3eqtr4g 2798	1 ⊢ (𝜑 → {⟨𝑥, 𝑦⟩ ∣ 𝜓} = {⟨𝑥, 𝑧⟩ ∣ 𝜒})

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1535  ∃wex 1774  {cab 2710  ⟨cop 4636  {copab 5211

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1963  ax-7 2003  ax-8 2106  ax-9 2114  ax-ext 2704

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1087  df-tru 1538  df-fal 1548  df-ex 1775  df-sb 2061  df-clab 2711  df-cleq 2725  df-clel 2812  df-rab 3433  df-v 3479  df-dif 3966  df-un 3968  df-ss 3980  df-nul 4340  df-if 4531  df-sn 4631  df-pr 4633  df-op 4637  df-opab 5212

This theorem is referenced by: (None)