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Theorem ackbij1lem1 9976
Description: Lemma for ackbij2 9999. (Contributed by Stefan O'Rear, 18-Nov-2014.)
Assertion
Ref Expression
ackbij1lem1 𝐴𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵𝐴))

Proof of Theorem ackbij1lem1
StepHypRef Expression
1 df-suc 6272 . . . 4 suc 𝐴 = (𝐴 ∪ {𝐴})
21ineq2i 4143 . . 3 (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3 indi 4207 . . 3 (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
42, 3eqtri 2766 . 2 (𝐵 ∩ suc 𝐴) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
5 disjsn 4647 . . . . 5 ((𝐵 ∩ {𝐴}) = ∅ ↔ ¬ 𝐴𝐵)
65biimpri 227 . . . 4 𝐴𝐵 → (𝐵 ∩ {𝐴}) = ∅)
76uneq2d 4097 . . 3 𝐴𝐵 → ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵𝐴) ∪ ∅))
8 un0 4324 . . 3 ((𝐵𝐴) ∪ ∅) = (𝐵𝐴)
97, 8eqtrdi 2794 . 2 𝐴𝐵 → ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = (𝐵𝐴))
104, 9eqtrid 2790 1 𝐴𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4   = wceq 1539  wcel 2106  cun 3885  cin 3886  c0 4256  {csn 4561  suc csuc 6268
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-rab 3073  df-v 3434  df-dif 3890  df-un 3892  df-in 3894  df-nul 4257  df-sn 4562  df-suc 6272
This theorem is referenced by:  ackbij1lem15  9990  ackbij1lem16  9991
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