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Theorem ackbij1lem1 9631
Description: Lemma for ackbij2 9654. (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 6165 . . . 4 suc 𝐴 = (𝐴 ∪ {𝐴})
21ineq2i 4136 . . 3 (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3 indi 4200 . . 3 (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
42, 3eqtri 2821 . 2 (𝐵 ∩ suc 𝐴) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
5 disjsn 4607 . . . . 5 ((𝐵 ∩ {𝐴}) = ∅ ↔ ¬ 𝐴𝐵)
65biimpri 231 . . . 4 𝐴𝐵 → (𝐵 ∩ {𝐴}) = ∅)
76uneq2d 4090 . . 3 𝐴𝐵 → ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵𝐴) ∪ ∅))
8 un0 4298 . . 3 ((𝐵𝐴) ∪ ∅) = (𝐵𝐴)
97, 8eqtrdi 2849 . 2 𝐴𝐵 → ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = (𝐵𝐴))
104, 9syl5eq 2845 1 𝐴𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4   = wceq 1538  wcel 2111  cun 3879  cin 3880  c0 4243  {csn 4525  suc csuc 6161
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-ral 3111  df-rab 3115  df-v 3443  df-dif 3884  df-un 3886  df-in 3888  df-nul 4244  df-sn 4526  df-suc 6165
This theorem is referenced by:  ackbij1lem15  9645  ackbij1lem16  9646
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