Theorem ackbij1lem2 10258

Description: Lemma for ackbij2 10280. (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 6392	. . . 4 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
2	1	ineq2i 4225	. . 3 ⊢ (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3		indi 4290	. . 3 ⊢ (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
4		uncom 4168	. . 3 ⊢ ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴))
5	2, 3, 4	3eqtri 2767	. 2 ⊢ (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴))
6		snssi 4813	. . . 4 ⊢ (𝐴 ∈ 𝐵 → {𝐴} ⊆ 𝐵)
7		sseqin2 4231	. . . 4 ⊢ ({𝐴} ⊆ 𝐵 ↔ (𝐵 ∩ {𝐴}) = {𝐴})
8	6, 7	sylib 218	. . 3 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ∩ {𝐴}) = {𝐴})
9	8	uneq1d 4177	. 2 ⊢ (𝐴 ∈ 𝐵 → ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴)) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))
10	5, 9	eqtrid 2787	1 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))

Syntax hints:   → wi 4   = wceq 1537   ∈ wcel 2106   ∪ cun 3961   ∩ cin 3962   ⊆ wss 3963  {csn 4631  suc csuc 6388

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434  df-v 3480  df-un 3968  df-in 3970  df-ss 3980  df-sn 4632  df-suc 6392

This theorem is referenced by:  ackbij1lem15  10271  ackbij1lem16  10272