Theorem ackbij1lem1 10214

Description: Lemma for ackbij2 10237. (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 6370	. . . 4 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
2	1	ineq2i 4209	. . 3 ⊢ (𝐵 ∩ suc 𝐴) = (𝐵 ∩ (𝐴 ∪ {𝐴}))
3		indi 4273	. . 3 ⊢ (𝐵 ∩ (𝐴 ∪ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
4	2, 3	eqtri 2760	. 2 ⊢ (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
5		disjsn 4715	. . . . 5 ⊢ ((𝐵 ∩ {𝐴}) = ∅ ↔ ¬ 𝐴 ∈ 𝐵)
6	5	biimpri 227	. . . 4 ⊢ (¬ 𝐴 ∈ 𝐵 → (𝐵 ∩ {𝐴}) = ∅)
7	6	uneq2d 4163	. . 3 ⊢ (¬ 𝐴 ∈ 𝐵 → ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ ∅))
8		un0 4390	. . 3 ⊢ ((𝐵 ∩ 𝐴) ∪ ∅) = (𝐵 ∩ 𝐴)
9	7, 8	eqtrdi 2788	. 2 ⊢ (¬ 𝐴 ∈ 𝐵 → ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = (𝐵 ∩ 𝐴))
10	4, 9	eqtrid 2784	1 ⊢ (¬ 𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵 ∩ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1541   ∈ wcel 2106   ∪ cun 3946   ∩ cin 3947  ∅c0 4322  {csn 4628  suc csuc 6366

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2703

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-fal 1554  df-ex 1782  df-sb 2068  df-clab 2710  df-cleq 2724  df-clel 2810  df-ral 3062  df-rab 3433  df-v 3476  df-dif 3951  df-un 3953  df-in 3955  df-nul 4323  df-sn 4629  df-suc 6370

This theorem is referenced by:  ackbij1lem15  10228  ackbij1lem16  10229