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Theorem riinint 4808
 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 4075 . . . . . . 7 ((𝑆𝑋𝑋𝑉) → 𝑆 ∈ V)
21expcom 115 . . . . . 6 (𝑋𝑉 → (𝑆𝑋𝑆 ∈ V))
32ralimdv 2503 . . . . 5 (𝑋𝑉 → (∀𝑘𝐼 𝑆𝑋 → ∀𝑘𝐼 𝑆 ∈ V))
43imp 123 . . . 4 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ∀𝑘𝐼 𝑆 ∈ V)
5 dfiin3g 4805 . . . 4 (∀𝑘𝐼 𝑆 ∈ V → 𝑘𝐼 𝑆 = ran (𝑘𝐼𝑆))
64, 5syl 14 . . 3 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → 𝑘𝐼 𝑆 = ran (𝑘𝐼𝑆))
76ineq2d 3282 . 2 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → (𝑋 𝑘𝐼 𝑆) = (𝑋 ran (𝑘𝐼𝑆)))
8 intun 3810 . . 3 ({𝑋} ∪ ran (𝑘𝐼𝑆)) = ( {𝑋} ∩ ran (𝑘𝐼𝑆))
9 intsng 3813 . . . . 5 (𝑋𝑉 {𝑋} = 𝑋)
109adantr 274 . . . 4 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → {𝑋} = 𝑋)
1110ineq1d 3281 . . 3 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ( {𝑋} ∩ ran (𝑘𝐼𝑆)) = (𝑋 ran (𝑘𝐼𝑆)))
128, 11syl5eq 2185 . 2 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ({𝑋} ∪ ran (𝑘𝐼𝑆)) = (𝑋 ran (𝑘𝐼𝑆)))
137, 12eqtr4d 2176 1 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → (𝑋 𝑘𝐼 𝑆) = ({𝑋} ∪ ran (𝑘𝐼𝑆)))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 103   = wceq 1332   ∈ wcel 1481  ∀wral 2417  Vcvv 2689   ∪ cun 3074   ∩ cin 3075   ⊆ wss 3076  {csn 3532  ∩ cint 3779  ∩ ciin 3822   ↦ cmpt 3997  ran crn 4548 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-14 1493  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122  ax-sep 4054  ax-pow 4106  ax-pr 4139 This theorem depends on definitions:  df-bi 116  df-3an 965  df-tru 1335  df-nf 1438  df-sb 1737  df-eu 2003  df-mo 2004  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-ral 2422  df-rex 2423  df-v 2691  df-un 3080  df-in 3082  df-ss 3089  df-pw 3517  df-sn 3538  df-pr 3539  df-op 3541  df-int 3780  df-iin 3824  df-br 3938  df-opab 3998  df-mpt 3999  df-cnv 4555  df-dm 4557  df-rn 4558 This theorem is referenced by: (None)
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