MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  ackbij1lem1 Structured version   Visualization version   GIF version

Theorem ackbij1lem1 10164
Description: Lemma for ackbij2 10187. (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 6327 . . . 4 suc 𝐴 = (𝐴 ∪ {𝐴})
21ineq2i 4173 . . 3 (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3 indi 4237 . . 3 (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
42, 3eqtri 2761 . 2 (𝐵 ∩ suc 𝐴) = ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴}))
5 disjsn 4676 . . . . 5 ((𝐵 ∩ {𝐴}) = ∅ ↔ ¬ 𝐴𝐵)
65biimpri 227 . . . 4 𝐴𝐵 → (𝐵 ∩ {𝐴}) = ∅)
76uneq2d 4127 . . 3 𝐴𝐵 → ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵𝐴) ∪ ∅))
8 un0 4354 . . 3 ((𝐵𝐴) ∪ ∅) = (𝐵𝐴)
97, 8eqtrdi 2789 . 2 𝐴𝐵 → ((𝐵𝐴) ∪ (𝐵 ∩ {𝐴})) = (𝐵𝐴))
104, 9eqtrid 2785 1 𝐴𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4   = wceq 1542  wcel 2107  cun 3912  cin 3913  c0 4286  {csn 4590  suc csuc 6323
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-fal 1555  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-ral 3062  df-rab 3407  df-v 3449  df-dif 3917  df-un 3919  df-in 3921  df-nul 4287  df-sn 4591  df-suc 6327
This theorem is referenced by:  ackbij1lem15  10178  ackbij1lem16  10179
  Copyright terms: Public domain W3C validator