Theorem oprab4 6087

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 4753	. . 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 6071	1 ⊢ {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑)} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑)}

Syntax hints:   ∧ wa 104   = wceq 1395   ∈ wcel 2200  ⟨cop 3670   × cxp 4721  {coprab 6014

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 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-14 2203  ax-ext 2211  ax-sep 4205  ax-pow 4262  ax-pr 4297

This theorem depends on definitions:  df-bi 117  df-3an 1004  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ral 2513  df-rex 2514  df-v 2802  df-un 3202  df-in 3204  df-ss 3211  df-pw 3652  df-sn 3673  df-pr 3674  df-op 3676  df-opab 4149  df-xp 4729  df-oprab 6017

This theorem is referenced by: (None)