Theorem brelg 30949

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 5118	. . 3 ⊢ (𝑅 ⊆ (𝐶 × 𝐷) → (𝐴𝑅𝐵 → 𝐴(𝐶 × 𝐷)𝐵))
2	1	imp 407	. 2 ⊢ ((𝑅 ⊆ (𝐶 × 𝐷) ∧ 𝐴𝑅𝐵) → 𝐴(𝐶 × 𝐷)𝐵)
3		brxp 5636	. 2 ⊢ (𝐴(𝐶 × 𝐷)𝐵 ↔ (𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷))
4	2, 3	sylib 217	1 ⊢ ((𝑅 ⊆ (𝐶 × 𝐷) ∧ 𝐴𝑅𝐵) → (𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷))

Syntax hints:   → wi 4   ∧ wa 396   ∈ wcel 2106   ⊆ wss 3887   class class class wbr 5074   × cxp 5587

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709  ax-sep 5223  ax-nul 5230  ax-pr 5352

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-rex 3070  df-rab 3073  df-v 3434  df-dif 3890  df-un 3892  df-in 3894  df-ss 3904  df-nul 4257  df-if 4460  df-sn 4562  df-pr 4564  df-op 4568  df-br 5075  df-opab 5137  df-xp 5595

This theorem is referenced by:  fpwrelmap  31068