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Theorem ackbij1lem1 9839
Description: Lemma for ackbij2 9862. (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 6224 . . . 4 suc 𝐴 = (𝐴 ∪ {𝐴})
21ineq2i 4129 . . 3 (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3 indi 4193 . . 3 (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
42, 3eqtri 2765 . 2 (𝐵 ∩ suc 𝐴) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
5 disjsn 4632 . . . . 5 ((𝐵 ∩ {𝐴}) = ∅ ↔ ¬ 𝐴𝐵)
65biimpri 231 . . . 4 𝐴𝐵 → (𝐵 ∩ {𝐴}) = ∅)
76uneq2d 4082 . . 3 𝐴𝐵 → ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵𝐴) ∪ ∅))
8 un0 4310 . . 3 ((𝐵𝐴) ∪ ∅) = (𝐵𝐴)
97, 8eqtrdi 2794 . 2 𝐴𝐵 → ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = (𝐵𝐴))
104, 9eqtrid 2789 1 𝐴𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4   = wceq 1543  wcel 2110  cun 3869  cin 3870  c0 4242  {csn 4546  suc csuc 6220
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2016  ax-8 2112  ax-9 2120  ax-ext 2708
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-tru 1546  df-fal 1556  df-ex 1788  df-sb 2071  df-clab 2715  df-cleq 2729  df-clel 2816  df-ral 3066  df-rab 3070  df-v 3415  df-dif 3874  df-un 3876  df-in 3878  df-nul 4243  df-sn 4547  df-suc 6224
This theorem is referenced by:  ackbij1lem15  9853  ackbij1lem16  9854
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