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Theorem ackbij1lem2 10219
Description: Lemma for ackbij2 10241. (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 6370 . . . 4 suc 𝐴 = (𝐴 ∪ {𝐴})
21ineq2i 4170 . . 3 (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3 indi 4237 . . 3 (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
4 uncom 4112 . . 3 ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ {𝐴}) ∪ (𝐵𝐴))
52, 3, 43eqtri 2792 . 2 (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ {𝐴}) ∪ (𝐵𝐴))
6 snssi 4753 . . . 4 (𝐴𝐵 → {𝐴} ⊆ 𝐵)
7 sseqin2 4176 . . . 4 ({𝐴} ⊆ 𝐵 ↔ (𝐵 ∩ {𝐴}) = {𝐴})
86, 7sylib 221 . . 3 (𝐴𝐵 → (𝐵 ∩ {𝐴}) = {𝐴})
98uneq1d 4121 . 2 (𝐴𝐵 → ((𝐵 ∩ {𝐴}) ∪ (𝐵𝐴)) = ({𝐴} ∪ (𝐵𝐴)))
105, 9eqtrid 2812 1 (𝐴𝐵 → (𝐵 ∩ suc 𝐴) = ({𝐴} ∪ (𝐵𝐴)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4   = wceq 1570  wcel 2146  cun 3904  cin 3905  wss 3906  {csn 4591  suc csuc 6366
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-un 3911  df-in 3913  df-ss 3923  df-sn 4592  df-suc 6370
This theorem is used by:  ackbij1lem15  10232  ackbij1lem16  10233
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