Theorem mpt2eq3dva 5569

Description: Slightly more general equality inference for the maps to notation. (Contributed by NM, 17-Oct-2013.)

Hypothesis

Ref	Expression
mpt2eq3dva.1	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝐶 = 𝐷)

Assertion

Ref	Expression
mpt2eq3dva	⊢ (𝜑 → (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐶) = (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐷))

Distinct variable groups:   𝜑,𝑥   𝜑,𝑦

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

Proof of Theorem mpt2eq3dva

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		mpt2eq3dva.1	. . . . . 6 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝐶 = 𝐷)
2	1	3expb 1105	. . . . 5 ⊢ ((𝜑 ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) → 𝐶 = 𝐷)
3	2	eqeq2d 2051	. . . 4 ⊢ ((𝜑 ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) → (𝑧 = 𝐶 ↔ 𝑧 = 𝐷))
4	3	pm5.32da 425	. . 3 ⊢ (𝜑 → (((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐶) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐷)))
5	4	oprabbidv 5559	. 2 ⊢ (𝜑 → {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐶)} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐷)})
6		df-mpt2 5517	. 2 ⊢ (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐶) = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐶)}
7		df-mpt2 5517	. 2 ⊢ (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐷) = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐷)}
8	5, 6, 7	3eqtr4g 2097	1 ⊢ (𝜑 → (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐶) = (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐷))

Syntax hints:   → wi 4   ∧ wa 97   ∧ w3a 885   = wceq 1243   ∈ wcel 1393  {coprab 5513   ↦ cmpt2 5514

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-11 1397  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-3an 887  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-oprab 5516  df-mpt2 5517

This theorem is referenced by:  mpt2eq3ia  5570  ofeq  5714  fmpt2co  5837  iseqeq2  9215  iseqeq3  9216  iseqval  9220