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Theorem indval2 12332
Description: Alternate value of the indicator function generator. (Contributed by Thierry Arnoux, 2-Feb-2017.)
Assertion
Ref Expression
indval2 ((𝑂 ∈ 𝑉 ∧ 𝐴 ⊆ 𝑂) → ((𝟭‘𝑂)‘𝐴) = ((𝐴 × {1}) ∪ ((𝑂 ∖ 𝐴) × {0})))

Proof of Theorem indval2
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 dfmpt3 6678 . . . 4 (𝑥 ∈ 𝑂 ↦ if(𝑥 ∈ 𝐴, 1, 0)) = ∪ 𝑥 ∈ 𝑂 ({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)})
2 indval 12330 . . . 4 ((𝑂 ∈ 𝑉 ∧ 𝐴 ⊆ 𝑂) → ((𝟭‘𝑂)‘𝐴) = (𝑥 ∈ 𝑂 ↦ if(𝑥 ∈ 𝐴, 1, 0)))
3 undif 4437 . . . . . 6 (𝐴 ⊆ 𝑂 ↔ (𝐴 ∪ (𝑂 ∖ 𝐴)) = 𝑂)
43bilani 510 . . . . 5 ((𝑂 ∈ 𝑉 ∧ 𝐴 ⊆ 𝑂) → (𝐴 ∪ (𝑂 ∖ 𝐴)) = 𝑂)
54iuneq1d 4978 . . . 4 ((𝑂 ∈ 𝑉 ∧ 𝐴 ⊆ 𝑂) → ∪ 𝑥 ∈ (𝐴 ∪ (𝑂 ∖ 𝐴))({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}) = ∪ 𝑥 ∈ 𝑂 ({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}))
61, 2, 53eqtr4a 2821 . . 3 ((𝑂 ∈ 𝑉 ∧ 𝐴 ⊆ 𝑂) → ((𝟭‘𝑂)‘𝐴) = ∪ 𝑥 ∈ (𝐴 ∪ (𝑂 ∖ 𝐴))({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}))
7 iunxun 5053 . . 3 ∪ 𝑥 ∈ (𝐴 ∪ (𝑂 ∖ 𝐴))({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}) = (∪ 𝑥 ∈ 𝐴 ({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}) ∪ ∪ 𝑥 ∈ (𝑂 ∖ 𝐴)({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}))
86, 7eqtrdi 2811 . 2 ((𝑂 ∈ 𝑉 ∧ 𝐴 ⊆ 𝑂) → ((𝟭‘𝑂)‘𝐴) = (∪ 𝑥 ∈ 𝐴 ({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}) ∪ ∪ 𝑥 ∈ (𝑂 ∖ 𝐴)({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)})))
9 iftrue 4487 . . . . . . 7 (𝑥 ∈ 𝐴 → if(𝑥 ∈ 𝐴, 1, 0) = 1)
109sneqd 4592 . . . . . 6 (𝑥 ∈ 𝐴 → {if(𝑥 ∈ 𝐴, 1, 0)} = {1})
1110xpeq2d 5681 . . . . 5 (𝑥 ∈ 𝐴 → ({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}) = ({𝑥} × {1}))
1211iuneq2i 4972 . . . 4 ∪ 𝑥 ∈ 𝐴 ({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}) = ∪ 𝑥 ∈ 𝐴 ({𝑥} × {1})
13 iunxpconst 5724 . . . 4 ∪ 𝑥 ∈ 𝐴 ({𝑥} × {1}) = (𝐴 × {1})
1412, 13eqtri 2783 . . 3 ∪ 𝑥 ∈ 𝐴 ({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}) = (𝐴 × {1})
15 eldifn 4067 . . . . . . 7 (𝑥 ∈ (𝑂 ∖ 𝐴) → ¬ 𝑥 ∈ 𝐴)
16 iffalse 4490 . . . . . . . 8 (¬ 𝑥 ∈ 𝐴 → if(𝑥 ∈ 𝐴, 1, 0) = 0)
1716sneqd 4592 . . . . . . 7 (¬ 𝑥 ∈ 𝐴 → {if(𝑥 ∈ 𝐴, 1, 0)} = {0})
1815, 17syl 18 . . . . . 6 (𝑥 ∈ (𝑂 ∖ 𝐴) → {if(𝑥 ∈ 𝐴, 1, 0)} = {0})
1918xpeq2d 5681 . . . . 5 (𝑥 ∈ (𝑂 ∖ 𝐴) → ({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}) = ({𝑥} × {0}))
2019iuneq2i 4972 . . . 4 ∪ 𝑥 ∈ (𝑂 ∖ 𝐴)({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}) = ∪ 𝑥 ∈ (𝑂 ∖ 𝐴)({𝑥} × {0})
21 iunxpconst 5724 . . . 4 ∪ 𝑥 ∈ (𝑂 ∖ 𝐴)({𝑥} × {0}) = ((𝑂 ∖ 𝐴) × {0})
2220, 21eqtri 2783 . . 3 ∪ 𝑥 ∈ (𝑂 ∖ 𝐴)({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}) = ((𝑂 ∖ 𝐴) × {0})
2314, 22uneq12i 4164 . 2 (∪ 𝑥 ∈ 𝐴 ({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)}) ∪ ∪ 𝑥 ∈ (𝑂 ∖ 𝐴)({𝑥} × {if(𝑥 ∈ 𝐴, 1, 0)})) = ((𝐴 × {1}) ∪ ((𝑂 ∖ 𝐴) × {0}))
248, 23eqtrdi 2811 1 ((𝑂 ∈ 𝑉 ∧ 𝐴 ⊆ 𝑂) → ((𝟭‘𝑂)‘𝐴) = ((𝐴 × {1}) ∪ ((𝑂 ∖ 𝐴) × {0})))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   = wceq 1569   ∈ wcel 2144   ⊆ wss 3896   ∖ cdif 4044   ∪ cun 4148  ifcif 4481  {csn 4584  ∪ ciun 4950   ↦ cmpt 5185   × cxp 5649  ‘cfv 6543  0cc0 11207  1c1 11208  𝟭cind 12327
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732  ax-rep 5231  ax-sep 5248  ax-nul 5259  ax-pow 5327  ax-pr 5391
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-mo 2564  df-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-ral 3077  df-rex 3087  df-reu 3366  df-rab 3413  df-v 3452  df-sbc 3739  df-csb 3848  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-pw 4558  df-sn 4585  df-pr 4587  df-op 4835  df-uni 4867  df-iun 4952  df-br 5103  df-opab 5167  df-mpt 5186  df-id 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-co 5660  df-dm 5661  df-rn 5662  df-res 5663  df-ima 5664  df-iota 6497  df-fun 6545  df-fn 6546  df-f 6547  df-f1 6548  df-fo 6549  df-f1o 6550  df-fv 6551  df-ind 12328
This theorem is used by:  indconst0  12339  indconst1  12340  gsumind  33956
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