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Theorem ackbij1lem1 9155
Description: Lemma for ackbij2 9178. (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 5842 . . . 4 suc 𝐴 = (𝐴 ∪ {𝐴})
21ineq2i 3919 . . 3 (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3 indi 3981 . . 3 (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
42, 3eqtri 2746 . 2 (𝐵 ∩ suc 𝐴) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
5 disjsn 4353 . . . . 5 ((𝐵 ∩ {𝐴}) = ∅ ↔ ¬ 𝐴𝐵)
65biimpri 218 . . . 4 𝐴𝐵 → (𝐵 ∩ {𝐴}) = ∅)
76uneq2d 3875 . . 3 𝐴𝐵 → ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵𝐴) ∪ ∅))
8 un0 4075 . . 3 ((𝐵𝐴) ∪ ∅) = (𝐵𝐴)
97, 8syl6eq 2774 . 2 𝐴𝐵 → ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = (𝐵𝐴))
104, 9syl5eq 2770 1 𝐴𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4   = wceq 1596  wcel 2103  cun 3678  cin 3679  c0 4023  {csn 4285  suc csuc 5838
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1835  ax-4 1850  ax-5 1952  ax-6 2018  ax-7 2054  ax-9 2112  ax-10 2132  ax-11 2147  ax-12 2160  ax-13 2355  ax-ext 2704
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1599  df-ex 1818  df-nf 1823  df-sb 2011  df-clab 2711  df-cleq 2717  df-clel 2720  df-nfc 2855  df-ral 3019  df-v 3306  df-dif 3683  df-un 3685  df-in 3687  df-nul 4024  df-sn 4286  df-suc 5842
This theorem is referenced by:  ackbij1lem15  9169  ackbij1lem16  9170
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