Theorem mpoeq3dva 6095

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

Hypothesis

Ref	Expression
mpoeq3dva.1	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝐶 = 𝐷)

Assertion

Ref	Expression
mpoeq3dva	⊢ (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 mpoeq3dva

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		mpoeq3dva.1	. . . . . 6 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝐶 = 𝐷)
2	1	3expb 1231	. . . . 5 ⊢ ((𝜑 ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) → 𝐶 = 𝐷)
3	2	eqeq2d 2243	. . . 4 ⊢ ((𝜑 ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) → (𝑧 = 𝐶 ↔ 𝑧 = 𝐷))
4	3	pm5.32da 452	. . 3 ⊢ (𝜑 → (((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐶) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐷)))
5	4	oprabbidv 6085	. 2 ⊢ (𝜑 → {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐶)} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐷)})
6		df-mpo 6033	. 2 ⊢ (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐶) = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐶)}
7		df-mpo 6033	. 2 ⊢ (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐷) = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 = 𝐷)}
8	5, 6, 7	3eqtr4g 2289	1 ⊢ (𝜑 → (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐶) = (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐷))

Syntax hints:   → wi 4   ∧ wa 104   ∧ w3a 1005   = wceq 1398   ∈ wcel 2202  {coprab 6029   ∈ cmpo 6030

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-11 1555  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-oprab 6032  df-mpo 6033

This theorem is referenced by:  mpoeq3ia  6096  mpoeq3dv  6097  ofeq  6247  fmpoco  6390  mapxpen  7077  seqeq2  10776  seqeq3  10777  grpsubpropd2  13768  mulgpropdg  13831  cnmpt2t  15104  cnmpt22  15105  cnmptcom  15109