Theorem ackbij1lem1 10127

Description: Lemma for ackbij2 10150. (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 6321	. . . 4 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
2	1	ineq2i 4167	. . 3 ⊢ (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3		indi 4234	. . 3 ⊢ (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
4	2, 3	eqtri 2757	. 2 ⊢ (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
5		disjsn 4666	. . . . 5 ⊢ ((𝐵 ∩ {𝐴}) = ∅ ↔ ¬ 𝐴 ∈ 𝐵)
6	5	biimpri 228	. . . 4 ⊢ (¬ 𝐴 ∈ 𝐵 → (𝐵 ∩ {𝐴}) = ∅)
7	6	uneq2d 4118	. . 3 ⊢ (¬ 𝐴 ∈ 𝐵 → ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ ∅))
8		un0 4344	. . 3 ⊢ ((𝐵 ∩ 𝐴) ∪ ∅) = (𝐵 ∩ 𝐴)
9	7, 8	eqtrdi 2785	. 2 ⊢ (¬ 𝐴 ∈ 𝐵 → ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = (𝐵 ∩ 𝐴))
10	4, 9	eqtrid 2781	1 ⊢ (¬ 𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵 ∩ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1541   ∈ wcel 2113   ∪ cun 3897   ∩ cin 3898  ∅c0 4283  {csn 4578  suc csuc 6317

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 2115  ax-9 2123  ax-ext 2706

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 2713  df-cleq 2726  df-clel 2809  df-ral 3050  df-rab 3398  df-v 3440  df-dif 3902  df-un 3904  df-in 3906  df-nul 4284  df-sn 4579  df-suc 6321

This theorem is referenced by:  ackbij1lem15  10141  ackbij1lem16  10142