Theorem cbvmptdavw 36624

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 2845	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
2	1	adantl 485	. . . 4 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
3		cbvmptdavw.1	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → 𝐵 = 𝐶)
4	3	eqeq2d 2773	. . . 4 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑡 = 𝐵 ↔ 𝑡 = 𝐶))
5	2, 4	anbi12d 641	. . 3 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → ((𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵) ↔ (𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶)))
6	5	cbvopab1davw 36621	. 2 ⊢ (𝜑 → {⟨𝑥, 𝑡⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵)} = {⟨𝑦, 𝑡⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶)})
7		df-mpt 5182	. 2 ⊢ (𝑥 ∈ 𝐴 ↦ 𝐵) = {⟨𝑥, 𝑡⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵)}
8		df-mpt 5182	. 2 ⊢ (𝑦 ∈ 𝐴 ↦ 𝐶) = {⟨𝑦, 𝑡⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶)}
9	6, 7, 8	3eqtr4g 2822	1 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↦ 𝐵) = (𝑦 ∈ 𝐴 ↦ 𝐶))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 399   = wceq 1560   ∈ wcel 2142  {copab 5162   ↦ cmpt 5181

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-ext 2734

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1100  df-tru 1563  df-fal 1573  df-ex 1800  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3456  df-dif 3907  df-un 3909  df-ss 3921  df-nul 4286  df-if 4481  df-sn 4583  df-pr 4585  df-op 4589  df-opab 5163  df-mpt 5182

This theorem is referenced by:  cbvproddavw  36637  cbvitgdavw  36638  cbvproddavw2  36653  cbvitgdavw2  36654