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Theorem riinint 5866
Description: Express a relative indexed intersection as an intersection. (Contributed by Stefan O'Rear, 22-Feb-2015.)
Assertion
Ref Expression
riinint ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → (𝑋 𝑘𝐼 𝑆) = ({𝑋} ∪ ran (𝑘𝐼𝑆)))
Distinct variable groups:   𝑘,𝑉   𝑘,𝑋
Allowed substitution hints:   𝑆(𝑘)   𝐼(𝑘)

Proof of Theorem riinint
StepHypRef Expression
1 ssexg 5242 . . . . . . 7 ((𝑆𝑋𝑋𝑉) → 𝑆 ∈ V)
21expcom 413 . . . . . 6 (𝑋𝑉 → (𝑆𝑋𝑆 ∈ V))
32ralimdv 3103 . . . . 5 (𝑋𝑉 → (∀𝑘𝐼 𝑆𝑋 → ∀𝑘𝐼 𝑆 ∈ V))
43imp 406 . . . 4 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ∀𝑘𝐼 𝑆 ∈ V)
5 dfiin3g 5863 . . . 4 (∀𝑘𝐼 𝑆 ∈ V → 𝑘𝐼 𝑆 = ran (𝑘𝐼𝑆))
64, 5syl 17 . . 3 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → 𝑘𝐼 𝑆 = ran (𝑘𝐼𝑆))
76ineq2d 4143 . 2 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → (𝑋 𝑘𝐼 𝑆) = (𝑋 ran (𝑘𝐼𝑆)))
8 intun 4908 . . 3 ({𝑋} ∪ ran (𝑘𝐼𝑆)) = ( {𝑋} ∩ ran (𝑘𝐼𝑆))
9 intsng 4913 . . . . 5 (𝑋𝑉 {𝑋} = 𝑋)
109adantr 480 . . . 4 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → {𝑋} = 𝑋)
1110ineq1d 4142 . . 3 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ( {𝑋} ∩ ran (𝑘𝐼𝑆)) = (𝑋 ran (𝑘𝐼𝑆)))
128, 11eqtrid 2790 . 2 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ({𝑋} ∪ ran (𝑘𝐼𝑆)) = (𝑋 ran (𝑘𝐼𝑆)))
137, 12eqtr4d 2781 1 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → (𝑋 𝑘𝐼 𝑆) = ({𝑋} ∪ ran (𝑘𝐼𝑆)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1539  wcel 2108  wral 3063  Vcvv 3422  cun 3881  cin 3882  wss 3883  {csn 4558   cint 4876   ciin 4922  cmpt 5153  ran crn 5581
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-mo 2540  df-eu 2569  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-int 4877  df-iin 4924  df-br 5071  df-opab 5133  df-mpt 5154  df-cnv 5588  df-dm 5590  df-rn 5591
This theorem is referenced by:  cmpfiiin  40435
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