Theorem brelg 30850

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 5114	. . 3 ⊢ (𝑅 ⊆ (𝐶 × 𝐷) → (𝐴𝑅𝐵 → 𝐴(𝐶 × 𝐷)𝐵))
2	1	imp 406	. 2 ⊢ ((𝑅 ⊆ (𝐶 × 𝐷) ∧ 𝐴𝑅𝐵) → 𝐴(𝐶 × 𝐷)𝐵)
3		brxp 5627	. 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 395   ∈ wcel 2108   ⊆ wss 3883   class class class wbr 5070   × cxp 5578

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-br 5071  df-opab 5133  df-xp 5586

This theorem is referenced by:  fpwrelmap  30970