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Theorem brelg 32629
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) → (𝐴𝐶𝐵𝐷))

Proof of Theorem brelg
StepHypRef Expression
1 ssbr 5192 . . 3 (𝑅 ⊆ (𝐶 × 𝐷) → (𝐴𝑅𝐵𝐴(𝐶 × 𝐷)𝐵))
21imp 406 . 2 ((𝑅 ⊆ (𝐶 × 𝐷) ∧ 𝐴𝑅𝐵) → 𝐴(𝐶 × 𝐷)𝐵)
3 brxp 5738 . 2 (𝐴(𝐶 × 𝐷)𝐵 ↔ (𝐴𝐶𝐵𝐷))
42, 3sylib 218 1 ((𝑅 ⊆ (𝐶 × 𝐷) ∧ 𝐴𝑅𝐵) → (𝐴𝐶𝐵𝐷))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395  wcel 2106  wss 3963   class class class wbr 5148   × cxp 5687
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706  ax-sep 5302  ax-nul 5312  ax-pr 5438
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ral 3060  df-rex 3069  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-br 5149  df-opab 5211  df-xp 5695
This theorem is referenced by:  fpwrelmap  32751
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