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Theorem riinint 5964
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 5292 . . . . . . 7 ((𝑆𝑋𝑋𝑉) → 𝑆 ∈ V)
21expcom 419 . . . . . 6 (𝑋𝑉 → (𝑆𝑋𝑆 ∈ V))
32ralimdv 3181 . . . . 5 (𝑋𝑉 → (∀𝑘𝐼 𝑆𝑋 → ∀𝑘𝐼 𝑆 ∈ V))
43imp 412 . . . 4 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ∀𝑘𝐼 𝑆 ∈ V)
5 dfiin3g 5961 . . . 4 (∀𝑘𝐼 𝑆 ∈ V → 𝑘𝐼 𝑆 = ran (𝑘𝐼𝑆))
64, 5syl 18 . . 3 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → 𝑘𝐼 𝑆 = ran (𝑘𝐼𝑆))
76ineq2d 4173 . 2 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → (𝑋 𝑘𝐼 𝑆) = (𝑋 ran (𝑘𝐼𝑆)))
8 intun 4947 . . 3 ({𝑋} ∪ ran (𝑘𝐼𝑆)) = ( {𝑋} ∩ ran (𝑘𝐼𝑆))
9 intsng 4950 . . . . 5 (𝑋𝑉 {𝑋} = 𝑋)
109adantr 486 . . . 4 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → {𝑋} = 𝑋)
1110ineq1d 4172 . . 3 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ( {𝑋} ∩ ran (𝑘𝐼𝑆)) = (𝑋 ran (𝑘𝐼𝑆)))
128, 11eqtrid 2812 . 2 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ({𝑋} ∪ ran (𝑘𝐼𝑆)) = (𝑋 ran (𝑘𝐼𝑆)))
137, 12eqtr4d 2803 1 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → (𝑋 𝑘𝐼 𝑆) = ({𝑋} ∪ ran (𝑘𝐼𝑆)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wa 401   = wceq 1570  wcel 2146  wral 3081  Vcvv 3457  cun 3904  cin 3905  wss 3906  {csn 4591   cint 4914   ciin 4959  cmpt 5194  ran crn 5664
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-10 2179  ax-11 2195  ax-12 2216  ax-ext 2737  ax-sep 5259  ax-pr 5406
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-nf 1817  df-sb 2100  df-mo 2569  df-eu 2599  df-clab 2744  df-cleq 2757  df-clel 2840  df-nfc 2914  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-int 4915  df-iin 4961  df-br 5112  df-opab 5176  df-mpt 5195  df-cnv 5671  df-dm 5673  df-rn 5674
This theorem is used by:  cmpfiiin  43488
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