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Theorem riinint 5489
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 4912 . . . . . . 7 ((𝑆𝑋𝑋𝑉) → 𝑆 ∈ V)
21expcom 450 . . . . . 6 (𝑋𝑉 → (𝑆𝑋𝑆 ∈ V))
32ralimdv 3065 . . . . 5 (𝑋𝑉 → (∀𝑘𝐼 𝑆𝑋 → ∀𝑘𝐼 𝑆 ∈ V))
43imp 444 . . . 4 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ∀𝑘𝐼 𝑆 ∈ V)
5 dfiin3g 5486 . . . 4 (∀𝑘𝐼 𝑆 ∈ V → 𝑘𝐼 𝑆 = ran (𝑘𝐼𝑆))
64, 5syl 17 . . 3 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → 𝑘𝐼 𝑆 = ran (𝑘𝐼𝑆))
76ineq2d 3922 . 2 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → (𝑋 𝑘𝐼 𝑆) = (𝑋 ran (𝑘𝐼𝑆)))
8 intun 4617 . . 3 ({𝑋} ∪ ran (𝑘𝐼𝑆)) = ( {𝑋} ∩ ran (𝑘𝐼𝑆))
9 intsng 4620 . . . . 5 (𝑋𝑉 {𝑋} = 𝑋)
109adantr 472 . . . 4 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → {𝑋} = 𝑋)
1110ineq1d 3921 . . 3 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ( {𝑋} ∩ ran (𝑘𝐼𝑆)) = (𝑋 ran (𝑘𝐼𝑆)))
128, 11syl5eq 2770 . 2 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ({𝑋} ∪ ran (𝑘𝐼𝑆)) = (𝑋 ran (𝑘𝐼𝑆)))
137, 12eqtr4d 2761 1 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → (𝑋 𝑘𝐼 𝑆) = ({𝑋} ∪ ran (𝑘𝐼𝑆)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 383   = wceq 1596  wcel 2103  wral 3014  Vcvv 3304  cun 3678  cin 3679  wss 3680  {csn 4285   cint 4583   ciin 4629  cmpt 4837  ran crn 5219
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1835  ax-4 1850  ax-5 1952  ax-6 2018  ax-7 2054  ax-9 2112  ax-10 2132  ax-11 2147  ax-12 2160  ax-13 2355  ax-ext 2704  ax-sep 4889  ax-nul 4897  ax-pr 5011
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1074  df-tru 1599  df-ex 1818  df-nf 1823  df-sb 2011  df-eu 2575  df-mo 2576  df-clab 2711  df-cleq 2717  df-clel 2720  df-nfc 2855  df-ral 3019  df-rex 3020  df-rab 3023  df-v 3306  df-dif 3683  df-un 3685  df-in 3687  df-ss 3694  df-nul 4024  df-if 4195  df-sn 4286  df-pr 4288  df-op 4292  df-int 4584  df-iin 4631  df-br 4761  df-opab 4821  df-mpt 4838  df-cnv 5226  df-dm 5228  df-rn 5229
This theorem is referenced by:  cmpfiiin  37679
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