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Theorem ackbij1lem2 10165
Description: Lemma for ackbij2 10187. (Contributed by Stefan O'Rear, 18-Nov-2014.)
Assertion
Ref Expression
ackbij1lem2 (𝐴𝐵 → (𝐵 ∩ suc 𝐴) = ({𝐴} ∪ (𝐵𝐴)))

Proof of Theorem ackbij1lem2
StepHypRef Expression
1 df-suc 6327 . . . 4 suc 𝐴 = (𝐴 ∪ {𝐴})
21ineq2i 4173 . . 3 (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3 indi 4237 . . 3 (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
4 uncom 4117 . . 3 ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ {𝐴}) ∪ (𝐵𝐴))
52, 3, 43eqtri 2765 . 2 (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ {𝐴}) ∪ (𝐵𝐴))
6 snssi 4772 . . . 4 (𝐴𝐵 → {𝐴} ⊆ 𝐵)
7 sseqin2 4179 . . . 4 ({𝐴} ⊆ 𝐵 ↔ (𝐵 ∩ {𝐴}) = {𝐴})
86, 7sylib 217 . . 3 (𝐴𝐵 → (𝐵 ∩ {𝐴}) = {𝐴})
98uneq1d 4126 . 2 (𝐴𝐵 → ((𝐵 ∩ {𝐴}) ∪ (𝐵𝐴)) = ({𝐴} ∪ (𝐵𝐴)))
105, 9eqtrid 2785 1 (𝐴𝐵 → (𝐵 ∩ suc 𝐴) = ({𝐴} ∪ (𝐵𝐴)))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1542  wcel 2107  cun 3912  cin 3913  wss 3914  {csn 4590  suc csuc 6323
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 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-rab 3407  df-v 3449  df-un 3919  df-in 3921  df-ss 3931  df-sn 4591  df-suc 6327
This theorem is referenced by:  ackbij1lem15  10178  ackbij1lem16  10179
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