Theorem ackbij1lem1 10110

Description: Lemma for ackbij2 10133. (Contributed by Stefan O'Rear, 18-Nov-2014.)

Assertion

Ref	Expression
ackbij1lem1	⊢ (¬ 𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵 ∩ 𝐴))

Proof of Theorem ackbij1lem1

Step	Hyp	Ref	Expression
1		df-suc 6312	. . . 4 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
2	1	ineq2i 4164	. . 3 ⊢ (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3		indi 4231	. . 3 ⊢ (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
4	2, 3	eqtri 2754	. 2 ⊢ (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
5		disjsn 4661	. . . . 5 ⊢ ((𝐵 ∩ {𝐴}) = ∅ ↔ ¬ 𝐴 ∈ 𝐵)
6	5	biimpri 228	. . . 4 ⊢ (¬ 𝐴 ∈ 𝐵 → (𝐵 ∩ {𝐴}) = ∅)
7	6	uneq2d 4115	. . 3 ⊢ (¬ 𝐴 ∈ 𝐵 → ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ ∅))
8		un0 4341	. . 3 ⊢ ((𝐵 ∩ 𝐴) ∪ ∅) = (𝐵 ∩ 𝐴)
9	7, 8	eqtrdi 2782	. 2 ⊢ (¬ 𝐴 ∈ 𝐵 → ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = (𝐵 ∩ 𝐴))
10	4, 9	eqtrid 2778	1 ⊢ (¬ 𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵 ∩ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1541   ∈ wcel 2111   ∪ cun 3895   ∩ cin 3896  ∅c0 4280  {csn 4573  suc csuc 6308

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ral 3048  df-rab 3396  df-v 3438  df-dif 3900  df-un 3902  df-in 3904  df-nul 4281  df-sn 4574  df-suc 6312

This theorem is referenced by:  ackbij1lem15  10124  ackbij1lem16  10125