Theorem mpt2eq3dva 5589

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 1139	. . . . 5 ⊢ ((𝜑 ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) → 𝐶 = 𝐷)
3	2	eqeq2d 2092	. . . 4 ⊢ ((𝜑 ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) → (𝑧 = 𝐶 ↔ 𝑧 = 𝐷))
4	3	pm5.32da 439	. . 3 ⊢ (𝜑 → (((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐶) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐷)))
5	4	oprabbidv 5579	. 2 ⊢ (𝜑 → {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐶)} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐷)})
6		df-mpt2 5537	. 2 ⊢ (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐶) = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐶)}
7		df-mpt2 5537	. 2 ⊢ (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐷) = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐷)}
8	5, 6, 7	3eqtr4g 2138	1 ⊢ (𝜑 → (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐶) = (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐷))

Syntax hints:   → wi 4   ∧ wa 102   ∧ w3a 919   = wceq 1284   ∈ wcel 1433  {coprab 5533   ↦ cmpt2 5534

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-3an 921  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-oprab 5536  df-mpt2 5537

This theorem is referenced by:  mpt2eq3ia  5590  ofeq  5734  fmpt2co  5857  iseqeq2  9435  iseqeq3  9436  iseqval  9440