Theorem brelg 32963

Description: Two things in a binary relation belong to the relation's domain. (Contributed by Thierry Arnoux, 29-Aug-2017.)

Assertion

Ref	Expression
brelg	⊢ ((𝑅 ⊆ (𝐶 × 𝐷) ∧ 𝐴𝑅𝐵) → (𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷))

Proof of Theorem brelg

Step	Hyp	Ref	Expression
1		ssbr 5154	. . 3 ⊢ (𝑅 ⊆ (𝐶 × 𝐷) → (𝐴𝑅𝐵 → 𝐴(𝐶 × 𝐷)𝐵))
2	1	imp 411	. 2 ⊢ ((𝑅 ⊆ (𝐶 × 𝐷) ∧ 𝐴𝑅𝐵) → 𝐴(𝐶 × 𝐷)𝐵)
3		brxp 5709	. 2 ⊢ (𝐴(𝐶 × 𝐷)𝐵 ↔ (𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷))
4	2, 3	sylib 221	1 ⊢ ((𝑅 ⊆ (𝐶 × 𝐷) ∧ 𝐴𝑅𝐵) → (𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷))

This proof depends on syntax axioms:   → wi 4   ∧ wa 400   ∈ wcel 2142   ⊆ wss 3904   class class class wbr 5108   × cxp 5658

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734  ax-sep 5256  ax-pr 5403

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1104  df-tru 1572  df-fal 1582  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3079  df-rex 3089  df-rab 3416  df-v 3456  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4487  df-sn 4589  df-pr 4591  df-op 4595  df-br 5109  df-opab 5173  df-xp 5666

This theorem is used by:  fpwrelmap  33089