Theorem ackbij1lem1 10233

Description: Lemma for ackbij2 10256. (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 6358	. . . 4 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
2	1	ineq2i 4192	. . 3 ⊢ (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3		indi 4259	. . 3 ⊢ (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
4	2, 3	eqtri 2758	. 2 ⊢ (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
5		disjsn 4687	. . . . 5 ⊢ ((𝐵 ∩ {𝐴}) = ∅ ↔ ¬ 𝐴 ∈ 𝐵)
6	5	biimpri 228	. . . 4 ⊢ (¬ 𝐴 ∈ 𝐵 → (𝐵 ∩ {𝐴}) = ∅)
7	6	uneq2d 4143	. . 3 ⊢ (¬ 𝐴 ∈ 𝐵 → ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ ∅))
8		un0 4369	. . 3 ⊢ ((𝐵 ∩ 𝐴) ∪ ∅) = (𝐵 ∩ 𝐴)
9	7, 8	eqtrdi 2786	. 2 ⊢ (¬ 𝐴 ∈ 𝐵 → ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = (𝐵 ∩ 𝐴))
10	4, 9	eqtrid 2782	1 ⊢ (¬ 𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵 ∩ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1540   ∈ wcel 2108   ∪ cun 3924   ∩ cin 3925  ∅c0 4308  {csn 4601  suc csuc 6354

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-ral 3052  df-rab 3416  df-v 3461  df-dif 3929  df-un 3931  df-in 3933  df-nul 4309  df-sn 4602  df-suc 6358

This theorem is referenced by:  ackbij1lem15  10247  ackbij1lem16  10248