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Theorem riinint 4768
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 4035 . . . . . . 7 ((𝑆𝑋𝑋𝑉) → 𝑆 ∈ V)
21expcom 115 . . . . . 6 (𝑋𝑉 → (𝑆𝑋𝑆 ∈ V))
32ralimdv 2475 . . . . 5 (𝑋𝑉 → (∀𝑘𝐼 𝑆𝑋 → ∀𝑘𝐼 𝑆 ∈ V))
43imp 123 . . . 4 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ∀𝑘𝐼 𝑆 ∈ V)
5 dfiin3g 4765 . . . 4 (∀𝑘𝐼 𝑆 ∈ V → 𝑘𝐼 𝑆 = ran (𝑘𝐼𝑆))
64, 5syl 14 . . 3 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → 𝑘𝐼 𝑆 = ran (𝑘𝐼𝑆))
76ineq2d 3245 . 2 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → (𝑋 𝑘𝐼 𝑆) = (𝑋 ran (𝑘𝐼𝑆)))
8 intun 3770 . . 3 ({𝑋} ∪ ran (𝑘𝐼𝑆)) = ( {𝑋} ∩ ran (𝑘𝐼𝑆))
9 intsng 3773 . . . . 5 (𝑋𝑉 {𝑋} = 𝑋)
109adantr 272 . . . 4 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → {𝑋} = 𝑋)
1110ineq1d 3244 . . 3 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ( {𝑋} ∩ ran (𝑘𝐼𝑆)) = (𝑋 ran (𝑘𝐼𝑆)))
128, 11syl5eq 2160 . 2 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → ({𝑋} ∪ ran (𝑘𝐼𝑆)) = (𝑋 ran (𝑘𝐼𝑆)))
137, 12eqtr4d 2151 1 ((𝑋𝑉 ∧ ∀𝑘𝐼 𝑆𝑋) → (𝑋 𝑘𝐼 𝑆) = ({𝑋} ∪ ran (𝑘𝐼𝑆)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103   = wceq 1314  wcel 1463  wral 2391  Vcvv 2658  cun 3037  cin 3038  wss 3039  {csn 3495   cint 3739   ciin 3782  cmpt 3957  ran crn 4508
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 681  ax-5 1406  ax-7 1407  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-8 1465  ax-10 1466  ax-11 1467  ax-i12 1468  ax-bndl 1469  ax-4 1470  ax-14 1475  ax-17 1489  ax-i9 1493  ax-ial 1497  ax-i5r 1498  ax-ext 2097  ax-sep 4014  ax-pow 4066  ax-pr 4099
This theorem depends on definitions:  df-bi 116  df-3an 947  df-tru 1317  df-nf 1420  df-sb 1719  df-eu 1978  df-mo 1979  df-clab 2102  df-cleq 2108  df-clel 2111  df-nfc 2245  df-ral 2396  df-rex 2397  df-v 2660  df-un 3043  df-in 3045  df-ss 3052  df-pw 3480  df-sn 3501  df-pr 3502  df-op 3504  df-int 3740  df-iin 3784  df-br 3898  df-opab 3958  df-mpt 3959  df-cnv 4515  df-dm 4517  df-rn 4518
This theorem is referenced by: (None)
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