Theorem relun 5800

Description: The union of two relations is a relation. Compare Exercise 5 of [TakeutiZaring] p. 25. (Contributed by NM, 12-Aug-1994.)

Assertion

Ref	Expression
relun	⊢ (Rel (𝐴 ∪ 𝐵) ↔ (Rel 𝐴 ∧ Rel 𝐵))

Proof of Theorem relun

Step	Hyp	Ref	Expression
1		unss 4143	. 2 ⊢ ((𝐴 ⊆ (V × V) ∧ 𝐵 ⊆ (V × V)) ↔ (𝐴 ∪ 𝐵) ⊆ (V × V))
2		df-rel 5670	. . 3 ⊢ (Rel 𝐴 ↔ 𝐴 ⊆ (V × V))
3		df-rel 5670	. . 3 ⊢ (Rel 𝐵 ↔ 𝐵 ⊆ (V × V))
4	2, 3	anbi12i 640	. 2 ⊢ ((Rel 𝐴 ∧ Rel 𝐵) ↔ (𝐴 ⊆ (V × V) ∧ 𝐵 ⊆ (V × V)))
5		df-rel 5670	. 2 ⊢ (Rel (𝐴 ∪ 𝐵) ↔ (𝐴 ∪ 𝐵) ⊆ (V × V))
6	1, 4, 5	3bitr4ri 307	1 ⊢ (Rel (𝐴 ∪ 𝐵) ↔ (Rel 𝐴 ∧ Rel 𝐵))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401  Vcvv 3457   ∪ cun 3904   ⊆ wss 3906   × cxp 5661  Rel wrel 5668

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 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-v 3459  df-un 3911  df-ss 3923  df-rel 5670

This theorem is used by:  difxp  6163  funun  6586  fununfun  6588  satfrel  35898  dfsucmap3  39172