Theorem oprab4 5993

Description: Two ways to state the domain of an operation. (Contributed by FL, 24-Jan-2010.)

Assertion

Ref	Expression
oprab4	⊢ {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑)} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑)}

Distinct variable group:   𝑥,𝑦,𝑧

Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧)   𝐴(𝑥,𝑦,𝑧)   𝐵(𝑥,𝑦,𝑧)

Proof of Theorem oprab4

Step	Hyp	Ref	Expression
1		opelxp 4693	. . 3 ⊢ (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))
2	1	anbi1i 458	. 2 ⊢ ((⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))
3	2	oprabbii 5977	1 ⊢ {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑)} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑)}

Syntax hints:   ∧ wa 104   = wceq 1364   ∈ wcel 2167  ⟨cop 3625   × cxp 4661  {coprab 5923

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-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-14 2170  ax-ext 2178  ax-sep 4151  ax-pow 4207  ax-pr 4242

This theorem depends on definitions:  df-bi 117  df-3an 982  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-ral 2480  df-rex 2481  df-v 2765  df-un 3161  df-in 3163  df-ss 3170  df-pw 3607  df-sn 3628  df-pr 3629  df-op 3631  df-opab 4095  df-xp 4669  df-oprab 5926

This theorem is referenced by: (None)