Theorem ackbij1lem2 10133

Description: Lemma for ackbij2 10155. (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 6316	. . . 4 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
2	1	ineq2i 4146	. . 3 ⊢ (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3		indi 4212	. . 3 ⊢ (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
4		uncom 4088	. . 3 ⊢ ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴))
5	2, 3, 4	3eqtri 2766	. 2 ⊢ (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴))
6		snssi 4717	. . . 4 ⊢ (𝐴 ∈ 𝐵 → {𝐴} ⊆ 𝐵)
7		sseqin2 4152	. . . 4 ⊢ ({𝐴} ⊆ 𝐵 ↔ (𝐵 ∩ {𝐴}) = {𝐴})
8	6, 7	sylib 219	. . 3 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ∩ {𝐴}) = {𝐴})
9	8	uneq1d 4097	. 2 ⊢ (𝐴 ∈ 𝐵 → ((𝐵 ∩ {𝐴}) ∪ (𝐵 ∩ 𝐴)) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))
10	5, 9	eqtrid 2786	1 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = ({𝐴} ∪ (𝐵 ∩ 𝐴)))

Syntax hints:   → wi 4   = wceq 1547   ∈ wcel 2119   ∪ cun 3881   ∩ cin 3882   ⊆ wss 3883  {csn 4555  suc csuc 6312

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-rab 3392  df-v 3433  df-un 3888  df-in 3890  df-ss 3900  df-sn 4556  df-suc 6316

This theorem is referenced by:  ackbij1lem15  10146  ackbij1lem16  10147