Theorem cbvopab2davw 37178

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 4840	. . . . . . . 8 ⊢ (𝑦 = 𝑧 → ⟨𝑥, 𝑦⟩ = ⟨𝑥, 𝑧⟩)
2	1	eqeq2d 2771	. . . . . . 7 ⊢ (𝑦 = 𝑧 → (𝑡 = ⟨𝑥, 𝑦⟩ ↔ 𝑡 = ⟨𝑥, 𝑧⟩))
3	2	adantl 487	. . . . . 6 ⊢ ((𝜑 ∧ 𝑦 = 𝑧) → (𝑡 = ⟨𝑥, 𝑦⟩ ↔ 𝑡 = ⟨𝑥, 𝑧⟩))
4		cbvopab2davw.1	. . . . . 6 ⊢ ((𝜑 ∧ 𝑦 = 𝑧) → (𝜓 ↔ 𝜒))
5	3, 4	anbi12d 644	. . . . 5 ⊢ ((𝜑 ∧ 𝑦 = 𝑧) → ((𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ (𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)))
6	5	cbvexdvaw 2071	. . . 4 ⊢ (𝜑 → (∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ ∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)))
7	6	exbidv 1953	. . 3 ⊢ (𝜑 → (∃𝑥∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ ∃𝑥∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)))
8	7	abbidv 2826	. 2 ⊢ (𝜑 → {𝑡 ∣ ∃𝑥∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)} = {𝑡 ∣ ∃𝑥∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)})
9		df-opab 5167	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜓} = {𝑡 ∣ ∃𝑥∃𝑦(𝑡 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)}
10		df-opab 5167	. 2 ⊢ {⟨𝑥, 𝑧⟩ ∣ 𝜒} = {𝑡 ∣ ∃𝑥∃𝑧(𝑡 = ⟨𝑥, 𝑧⟩ ∧ 𝜒)}
11	8, 9, 10	3eqtr4g 2820	1 ⊢ (𝜑 → {⟨𝑥, 𝑦⟩ ∣ 𝜓} = {⟨𝑥, 𝑧⟩ ∣ 𝜒})

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569  ∃wex 1811  {cab 2738  ⟨cop 4834  {copab 5166

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-opab 5167

This theorem is used by: (None)