Theorem brelg 33067

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 5153	. . 3 ⊢ (𝑅 ⊆ (𝐶 × 𝐷) → (𝐴𝑅𝐵 → 𝐴(𝐶 × 𝐷)𝐵))
2	1	imp 412	. 2 ⊢ ((𝑅 ⊆ (𝐶 × 𝐷) ∧ 𝐴𝑅𝐵) → 𝐴(𝐶 × 𝐷)𝐵)
3		brxp 5708	. 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 401   ∈ wcel 2145   ⊆ wss 3902   class class class wbr 5107   × cxp 5657

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734  ax-sep 5255  ax-pr 5402

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 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-br 5108  df-opab 5172  df-xp 5665

This theorem is used by:  fpwrelmap  33191