Theorem cbvmptdavw 36836

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 2848	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
2	1	adantl 487	. . . 4 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
3		cbvmptdavw.1	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → 𝐵 = 𝐶)
4	3	eqeq2d 2776	. . . 4 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑡 = 𝐵 ↔ 𝑡 = 𝐶))
5	2, 4	anbi12d 644	. . 3 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → ((𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵) ↔ (𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶)))
6	5	cbvopab1davw 36833	. 2 ⊢ (𝜑 → {⟨𝑥, 𝑡⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵)} = {⟨𝑦, 𝑡⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶)})
7		df-mpt 5195	. 2 ⊢ (𝑥 ∈ 𝐴 ↦ 𝐵) = {⟨𝑥, 𝑡⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵)}
8		df-mpt 5195	. 2 ⊢ (𝑦 ∈ 𝐴 ↦ 𝐶) = {⟨𝑦, 𝑡⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶)}
9	6, 7, 8	3eqtr4g 2825	1 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↦ 𝐵) = (𝑦 ∈ 𝐴 ↦ 𝐶))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2146  {copab 5175   ↦ cmpt 5194

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-opab 5176  df-mpt 5195

This theorem is used by:  cbvproddavw  36849  cbvitgdavw  36850  cbvproddavw2  36865  cbvitgdavw2  36866