Theorem ackbij1lem2 10180

Description: Lemma for ackbij2 10202. (Contributed by Stefan O'Rear, 18-Nov-2014.)

Assertion

Ref	Expression
ackbij1lem2	⊢ (𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))

Proof of Theorem ackbij1lem2

Step	Hyp	Ref	Expression
1		df-suc 6341	. . . 4 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
2	1	ineq2i 4183	. . 3 ⊢ (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3		indi 4250	. . 3 ⊢ (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
4		uncom 4124	. . 3 ⊢ ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴))
5	2, 3, 4	3eqtri 2757	. 2 ⊢ (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴))
6		snssi 4775	. . . 4 ⊢ (𝐴 ∈ 𝐵 → {𝐴} ⊆ 𝐵)
7		sseqin2 4189	. . . 4 ⊢ ({𝐴} ⊆ 𝐵 ↔ (𝐵 ∩ {𝐴}) = {𝐴})
8	6, 7	sylib 218	. . 3 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ∩ {𝐴}) = {𝐴})
9	8	uneq1d 4133	. 2 ⊢ (𝐴 ∈ 𝐵 → ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴)) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))
10	5, 9	eqtrid 2777	1 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))

Syntax hints:   → wi 4   = wceq 1540   ∈ wcel 2109   ∪ cun 3915   ∩ cin 3916   ⊆ wss 3917  {csn 4592  suc csuc 6337

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 2008  ax-8 2111  ax-9 2119  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-rab 3409  df-v 3452  df-un 3922  df-in 3924  df-ss 3934  df-sn 4593  df-suc 6341

This theorem is referenced by:  ackbij1lem15  10193  ackbij1lem16  10194