Theorem mpteq12da 5198

Description: An equality inference for the maps-to notation. (Contributed by Glauco Siliprandi, 23-Oct-2021.) Remove dependency on ax-10 2182. (Revised by SN, 11-Nov-2024.)

Hypotheses

Ref	Expression
mpteq12da.1	⊢ Ⅎ𝑥𝜑
mpteq12da.2	⊢ (𝜑 → 𝐴 = 𝐶)
mpteq12da.3	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐵 = 𝐷)

Assertion

Ref	Expression
mpteq12da	⊢ (𝜑 → (𝑥 ∈ 𝐴 ↦ 𝐵) = (𝑥 ∈ 𝐶 ↦ 𝐷))

Proof of Theorem mpteq12da

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		mpteq12da.1	. . 3 ⊢ Ⅎ𝑥𝜑
2		nfv 1941	. . 3 ⊢ Ⅎ𝑦𝜑
3		mpteq12da.3	. . . . . 6 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐵 = 𝐷)
4	3	eqeq2d 2780	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝑦 = 𝐵 ↔ 𝑦 = 𝐷))
5	4	pm5.32da 589	. . . 4 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐷)))
6		mpteq12da.2	. . . . . 6 ⊢ (𝜑 → 𝐴 = 𝐶)
7	6	eleq2d 2855	. . . . 5 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐶))
8	7	anbi1d 642	. . . 4 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐷) ↔ (𝑥 ∈ 𝐶 ∧ 𝑦 = 𝐷)))
9	5, 8	bitrd 282	. . 3 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ 𝑦 = 𝐷)))
10	1, 2, 9	opabbid 5180	. 2 ⊢ (𝜑 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 = 𝐷)})
11		df-mpt 5197	. 2 ⊢ (𝑥 ∈ 𝐴 ↦ 𝐵) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐵)}
12		df-mpt 5197	. 2 ⊢ (𝑥 ∈ 𝐶 ↦ 𝐷) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 = 𝐷)}
13	10, 11, 12	3eqtr4g 2829	1 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↦ 𝐵) = (𝑥 ∈ 𝐶 ↦ 𝐷))

Syntax hints:   → wi 4   ∧ wa 400   = wceq 1567  Ⅎwnf 1810   ∈ wcel 2149  {copab 5177   ↦ cmpt 5196

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-12 2219  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1807  df-nf 1811  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-opab 5178  df-mpt 5197

This theorem is referenced by:  mpteq12df  5199  mpteq2da  5207  smflimmpt  47415  smfsupmpt  47420  smfinfmpt  47424  smflimsupmpt  47434  smfliminfmpt  47437