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Theorem ackbij1lem2 10260
Description: Lemma for ackbij2 10282. (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 6390 . . . 4 suc 𝐴 = (𝐴 ∪ {𝐴})
21ineq2i 4217 . . 3 (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3 indi 4284 . . 3 (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
4 uncom 4158 . . 3 ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ {𝐴}) ∪ (𝐵𝐴))
52, 3, 43eqtri 2769 . 2 (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ {𝐴}) ∪ (𝐵𝐴))
6 snssi 4808 . . . 4 (𝐴𝐵 → {𝐴} ⊆ 𝐵)
7 sseqin2 4223 . . . 4 ({𝐴} ⊆ 𝐵 ↔ (𝐵 ∩ {𝐴}) = {𝐴})
86, 7sylib 218 . . 3 (𝐴𝐵 → (𝐵 ∩ {𝐴}) = {𝐴})
98uneq1d 4167 . 2 (𝐴𝐵 → ((𝐵 ∩ {𝐴}) ∪ (𝐵𝐴)) = ({𝐴} ∪ (𝐵𝐴)))
105, 9eqtrid 2789 1 (𝐴𝐵 → (𝐵 ∩ suc 𝐴) = ({𝐴} ∪ (𝐵𝐴)))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1540  wcel 2108  cun 3949  cin 3950  wss 3951  {csn 4626  suc csuc 6386
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-rab 3437  df-v 3482  df-un 3956  df-in 3958  df-ss 3968  df-sn 4627  df-suc 6390
This theorem is referenced by:  ackbij1lem15  10273  ackbij1lem16  10274
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