Theorem cbvopab2davw 36573

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 4826	. . . . . . . 8 ⊢ (𝑦 = 𝑧 → ⟨𝑥, 𝑦⟩ = ⟨𝑥, 𝑧⟩)
2	1	eqeq2d 2767	. . . . . . 7 ⊢ (𝑦 = 𝑧 → (𝑡 = ⟨𝑥, 𝑦⟩ ↔ 𝑡 = ⟨𝑥, 𝑧⟩))
3	2	adantl 484	. . . . . 6 ⊢ ((𝜑 ∧ 𝑦 = 𝑧) → (𝑡 = ⟨𝑥, 𝑦⟩ ↔ 𝑡 = ⟨𝑥, 𝑧⟩))
4		cbvopab2davw.1	. . . . . 6 ⊢ ((𝜑 ∧ 𝑦 = 𝑧) → (𝜓 ↔ 𝜒))
5	3, 4	anbi12d 640	. . . . 5 ⊢ ((𝜑 ∧ 𝑦 = 𝑧) → ((𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ (𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)))
6	5	cbvexdvaw 2053	. . . 4 ⊢ (𝜑 → (∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ ∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)))
7	6	exbidv 1935	. . 3 ⊢ (𝜑 → (∃𝑥∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ ∃𝑥∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)))
8	7	abbidv 2822	. 2 ⊢ (𝜑 → {𝑡 ∣ ∃𝑥∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)} = {𝑡 ∣ ∃𝑥∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)})
9		df-opab 5157	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜓} = {𝑡 ∣ ∃𝑥∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)}
10		df-opab 5157	. 2 ⊢ {⟨𝑥, 𝑧⟩ ∣ 𝜒} = {𝑡 ∣ ∃𝑥∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)}
11	8, 9, 10	3eqtr4g 2816	1 ⊢ (𝜑 → {⟨𝑥, 𝑦⟩ ∣ 𝜓} = {⟨𝑥, 𝑧⟩ ∣ 𝜒})

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398   = wceq 1554  ∃wex 1793  {cab 2734  ⟨cop 4582  {copab 5156

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1809  ax-4 1823  ax-5 1924  ax-6 1981  ax-7 2022  ax-8 2138  ax-9 2146  ax-ext 2728

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 857  df-3an 1097  df-tru 1557  df-fal 1567  df-ex 1794  df-sb 2085  df-clab 2735  df-cleq 2748  df-clel 2831  df-rab 3409  df-v 3450  df-dif 3902  df-un 3904  df-ss 3916  df-nul 4281  df-if 4475  df-sn 4577  df-pr 4579  df-op 4583  df-opab 5157

This theorem is referenced by: (None)