Theorem oprab4 5913

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 4634	. . 3 ⊢ (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))
2	1	anbi1i 454	. 2 ⊢ ((⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))
3	2	oprabbii 5897	1 ⊢ {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑)} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑)}

Syntax hints:   ∧ wa 103   = wceq 1343   ∈ wcel 2136  ⟨cop 3579   × cxp 4602  {coprab 5843

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-14 2139  ax-ext 2147  ax-sep 4100  ax-pow 4153  ax-pr 4187

This theorem depends on definitions:  df-bi 116  df-3an 970  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ral 2449  df-rex 2450  df-v 2728  df-un 3120  df-in 3122  df-ss 3129  df-pw 3561  df-sn 3582  df-pr 3583  df-op 3585  df-opab 4044  df-xp 4610  df-oprab 5846

This theorem is referenced by: (None)