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Theorem brelg 33270
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
StepHypRef Expression
1 ssbr 5152 . . 3 (𝑅 ⊆ (𝐶 × 𝐷) → (𝐴𝑅𝐵 → 𝐴(𝐶 × 𝐷)𝐵))
21imp 412 . 2 ((𝑅 ⊆ (𝐶 × 𝐷) ∧ 𝐴𝑅𝐵) → 𝐴(𝐶 × 𝐷)𝐵)
3 brxp 5704 . 2 (𝐴(𝐶 × 𝐷)𝐵 ↔ (𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷))
42, 3sylib 221 1 ((𝑅 ⊆ (𝐶 × 𝐷) ∧ 𝐴𝑅𝐵) → (𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∈ wcel 2147   ⊆ wss 3900   class class class wbr 5106   × cxp 5653
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5252  ax-pr 5395
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5171  df-xp 5661
This theorem is used by:  fpwrelmap  33392
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