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Theorem ackbij1lem2 10337
Description: Lemma for ackbij2 10359. (Contributed by Stefan O'Rear, 18-Nov-2014.)
Assertion
Ref Expression
ackbij1lem2 (𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))

Proof of Theorem ackbij1lem2
StepHypRef Expression
1 df-suc 6394 . . . 4 suc 𝐴 = (𝐴 ∪ {𝐴})
21ineq2i 4117 . . 3 (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3 indi 4232 . . 3 (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
4 uncom 4160 . . 3 ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴))
52, 3, 43eqtri 2790 . 2 (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴))
6 snssi 4748 . . . 4 (𝐴 ∈ 𝐵 → {𝐴} ⊆ 𝐵)
7 sseqin2 4123 . . . 4 ({𝐴} ⊆ 𝐵 ↔ (𝐵 ∩ {𝐴}) = {𝐴})
86, 7sylib 221 . . 3 (𝐴 ∈ 𝐵 → (𝐵 ∩ {𝐴}) = {𝐴})
98uneq1d 4169 . 2 (𝐴 ∈ 𝐵 → ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴)) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))
105, 9eqtrid 2810 1 (𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∈ wcel 2147   ⊆ wss 3900   ∩ cin 4092   ∪ cun 4152  {csn 4588  suc csuc 6390
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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735
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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-in 4094  df-un 4154  df-sn 4589  df-suc 6394
This theorem is used by:  ackbij1lem15  10350  ackbij1lem16  10351
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