Theorem cbvmptdavw 36495

Description: Change bound variable in a maps-to function. Deduction form. (Contributed by GG, 14-Aug-2025.)

Hypothesis

Ref	Expression
cbvmptdavw.1	⊢ ((𝜑 ∧ 𝑥 = 𝑦) → 𝐵 = 𝐶)

Assertion

Ref	Expression
cbvmptdavw	⊢ (𝜑 → (𝑥 ∈ 𝐴 ↦ 𝐵) = (𝑦 ∈ 𝐴 ↦ 𝐶))

Distinct variable groups:   𝜑,𝑥,𝑦   𝑥,𝐴,𝑦   𝑦,𝐵   𝑥,𝐶

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

Proof of Theorem cbvmptdavw

Dummy variable 𝑡 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq1w 2822	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
2	1	adantl 482	. . . 4 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
3		cbvmptdavw.1	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → 𝐵 = 𝐶)
4	3	eqeq2d 2750	. . . 4 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑡 = 𝐵 ↔ 𝑡 = 𝐶))
5	2, 4	anbi12d 638	. . 3 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → ((𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵) ↔ (𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶)))
6	5	cbvopab1davw 36492	. 2 ⊢ (𝜑 → {⟨𝑥, 𝑡⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵)} = {⟨𝑦, 𝑡⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶)})
7		df-mpt 5154	. 2 ⊢ (𝑥 ∈ 𝐴 ↦ 𝐵) = {⟨𝑥, 𝑡⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵)}
8		df-mpt 5154	. 2 ⊢ (𝑦 ∈ 𝐴 ↦ 𝐶) = {⟨𝑦, 𝑡⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶)}
9	6, 7, 8	3eqtr4g 2799	1 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↦ 𝐵) = (𝑦 ∈ 𝐴 ↦ 𝐶))

Syntax hints:   → wi 4   ↔ wb 207   ∧ wa 396   = wceq 1547   ∈ wcel 2119  {copab 5134   ↦ cmpt 5153

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-rab 3392  df-v 3433  df-dif 3886  df-un 3888  df-ss 3900  df-nul 4262  df-if 4455  df-sn 4556  df-pr 4558  df-op 4562  df-opab 5135  df-mpt 5154

This theorem is referenced by:  cbvproddavw  36508  cbvitgdavw  36509  cbvproddavw2  36524  cbvitgdavw2  36525