Theorem relfld 6301

Description: The double union of a relation is its field. (Contributed by NM, 17-Sep-2006.)

Assertion

Ref	Expression
relfld	⊢ (Rel 𝑅 → ∪ ∪ 𝑅 = (dom 𝑅 ∪ ran 𝑅))

Proof of Theorem relfld

Step	Hyp	Ref	Expression
1		relssdmrn 6295	. . . 4 ⊢ (Rel 𝑅 → 𝑅 ⊆ (dom 𝑅 × ran 𝑅))
2		uniss 4880	. . . 4 ⊢ (𝑅 ⊆ (dom 𝑅 × ran 𝑅) → ∪ 𝑅 ⊆ ∪ (dom 𝑅 × ran 𝑅))
3		uniss 4880	. . . 4 ⊢ (∪ 𝑅 ⊆ ∪ (dom 𝑅 × ran 𝑅) → ∪ ∪ 𝑅 ⊆ ∪ ∪ (dom 𝑅 × ran 𝑅))
4	1, 2, 3	3syl 19	. . 3 ⊢ (Rel 𝑅 → ∪ ∪ 𝑅 ⊆ ∪ ∪ (dom 𝑅 × ran 𝑅))
5		unixpss 5797	. . 3 ⊢ ∪ ∪ (dom 𝑅 × ran 𝑅) ⊆ (dom 𝑅 ∪ ran 𝑅)
6	4, 5	sstrdi 3925	. 2 ⊢ (Rel 𝑅 → ∪ ∪ 𝑅 ⊆ (dom 𝑅 ∪ ran 𝑅))
7		dmrnssfld 5968	. . 3 ⊢ (dom 𝑅 ∪ ran 𝑅) ⊆ ∪ ∪ 𝑅
8	7	a1i 11	. 2 ⊢ (Rel 𝑅 → (dom 𝑅 ∪ ran 𝑅) ⊆ ∪ ∪ 𝑅)
9	6, 8	eqssd 3930	1 ⊢ (Rel 𝑅 → ∪ ∪ 𝑅 = (dom 𝑅 ∪ ran 𝑅))

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ⊆ wss 3900   ∪ cun 4152  ∪ cuni 4872   × cxp 5656  dom cdm 5658  ran crn 5659  Rel wrel 5663

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 5255  ax-pr 5398

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-pw 4562  df-sn 4589  df-pr 4591  df-op 4841  df-uni 4873  df-br 5109  df-opab 5174  df-xp 5664  df-rel 5665  df-cnv 5666  df-dm 5668  df-rn 5669

This theorem is used by:  relresfld  6302  relresfldOLD  6303  unidmrn  6306  relcnvfld  6307  unixp  6309  relexp0  15219  relexpfld  15245  rtrclreclem4  15257  dfrtrcl2  15258  lefld  18816  fvmptiunrelexplb0da  44946