Theorem cbvmptvw2 37119

Description: Change bound variable and domain in a maps-to function, using implicit substitution. (Contributed by GG, 14-Aug-2025.)

Hypotheses

Ref	Expression
cbvmptvw2.1	⊢ (𝑥 = 𝑦 → 𝐶 = 𝐷)
cbvmptvw2.2	⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)

Assertion

Ref	Expression
cbvmptvw2	⊢ (𝑥 ∈ 𝐴 ↦ 𝐶) = (𝑦 ∈ 𝐵 ↦ 𝐷)

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

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

Proof of Theorem cbvmptvw2

Dummy variable 𝑡 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq1w 2843	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
2		cbvmptvw2.2	. . . . . 6 ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
3	2	eleq2d 2846	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
4	1, 3	bitrd 282	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
5		cbvmptvw2.1	. . . . 5 ⊢ (𝑥 = 𝑦 → 𝐶 = 𝐷)
6	5	eqeq2d 2771	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑡 = 𝐶 ↔ 𝑡 = 𝐷))
7	4, 6	anbi12d 644	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐶) ↔ (𝑦 ∈ 𝐵 ∧ 𝑡 = 𝐷)))
8	7	cbvopab1v 5182	. 2 ⊢ {⟨𝑥, 𝑡⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐶)} = {⟨𝑦, 𝑡⟩ ∣ (𝑦 ∈ 𝐵 ∧ 𝑡 = 𝐷)}
9		df-mpt 5186	. 2 ⊢ (𝑥 ∈ 𝐴 ↦ 𝐶) = {⟨𝑥, 𝑡⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐶)}
10		df-mpt 5186	. 2 ⊢ (𝑦 ∈ 𝐵 ↦ 𝐷) = {⟨𝑦, 𝑡⟩ ∣ (𝑦 ∈ 𝐵 ∧ 𝑡 = 𝐷)}
11	8, 9, 10	3eqtr4i 2793	1 ⊢ (𝑥 ∈ 𝐴 ↦ 𝐶) = (𝑦 ∈ 𝐵 ↦ 𝐷)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1569   ∈ wcel 2144  {copab 5166   ↦ cmpt 5185

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  df-mpt 5186

This theorem is used by: (None)