ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  oprab4 GIF version

Theorem oprab4 5939
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⟩ ∣ ((𝑥𝐴𝑦𝐵) ∧ 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
StepHypRef Expression
1 opelxp 4652 . . 3 (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ↔ (𝑥𝐴𝑦𝐵))
21anbi1i 458 . 2 ((⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑) ↔ ((𝑥𝐴𝑦𝐵) ∧ 𝜑))
32oprabbii 5923 1 {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑)} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝐴𝑦𝐵) ∧ 𝜑)}
Colors of variables: wff set class
Syntax hints:  wa 104   = wceq 1353  wcel 2148  cop 3594   × cxp 4620  {coprab 5869
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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-14 2151  ax-ext 2159  ax-sep 4118  ax-pow 4171  ax-pr 4205
This theorem depends on definitions:  df-bi 117  df-3an 980  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-rex 2461  df-v 2739  df-un 3133  df-in 3135  df-ss 3142  df-pw 3576  df-sn 3597  df-pr 3598  df-op 3600  df-opab 4062  df-xp 4628  df-oprab 5872
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator