Theorem ackbij1lem2 10111

Description: Lemma for ackbij2 10133. (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 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		uncom 4105	. . 3 ⊢ ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴))
5	2, 3, 4	3eqtri 2758	. 2 ⊢ (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴))
6		snssi 4757	. . . 4 ⊢ (𝐴 ∈ 𝐵 → {𝐴} ⊆ 𝐵)
7		sseqin2 4170	. . . 4 ⊢ ({𝐴} ⊆ 𝐵 ↔ (𝐵 ∩ {𝐴}) = {𝐴})
8	6, 7	sylib 218	. . 3 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ∩ {𝐴}) = {𝐴})
9	8	uneq1d 4114	. 2 ⊢ (𝐴 ∈ 𝐵 → ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴)) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))
10	5, 9	eqtrid 2778	1 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))

Syntax hints:   → wi 4   = wceq 1541   ∈ wcel 2111   ∪ cun 3895   ∩ cin 3896   ⊆ wss 3897  {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-3an 1088  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rab 3396  df-v 3438  df-un 3902  df-in 3904  df-ss 3914  df-sn 4574  df-suc 6312

This theorem is referenced by:  ackbij1lem15  10124  ackbij1lem16  10125