Theorem oprab4 5575

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 4374	. . 3 ⊢ (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))
2	1	anbi1i 431	. 2 ⊢ ((⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))
3	2	oprabbii 5560	1 ⊢ {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ (⟨𝑥, 𝑦⟩ ∈ (𝐴 × 𝐵) ∧ 𝜑)} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑)}

Syntax hints:   ∧ wa 97   = wceq 1243   ∈ wcel 1393  ⟨cop 3378   × cxp 4343  {coprab 5513

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-14 1405  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022  ax-sep 3875  ax-pow 3927  ax-pr 3944

This theorem depends on definitions:  df-bi 110  df-3an 887  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-ral 2311  df-rex 2312  df-v 2559  df-un 2922  df-in 2924  df-ss 2931  df-pw 3361  df-sn 3381  df-pr 3382  df-op 3384  df-opab 3819  df-xp 4351  df-oprab 5516

This theorem is referenced by: (None)