Theorem oprab4 6159

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 4804	. . 3 ⊢ (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))
2	1	anbi1i 462	. 2 ⊢ ((⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))
3	2	oprabbii 6143	1 ⊢ {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑)} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑)}

This proof depends on syntax axioms:   ∧ wa 104   = wceq 1402   ∈ wcel 2209  ⟨cop 3712   × cxp 4772  {coprab 6086

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-sep 4249  ax-pow 4311  ax-pr 4346

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-v 2823  df-un 3224  df-in 3226  df-ss 3233  df-pw 3690  df-sn 3715  df-pr 3716  df-op 3718  df-opab 4193  df-xp 4780  df-oprab 6089

This theorem is used by: (None)