Theorem ackbij1lem1 10132

Description: Lemma for ackbij2 10155. (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 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	2, 3	eqtri 2762	. 2 ⊢ (𝐵 ∩ suc 𝐴) = ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴}))
5		disjsn 4643	. . . . 5 ⊢ ((𝐵 ∩ {𝐴}) = ∅ ↔ ¬ 𝐴 ∈ 𝐵)
6	5	biimpri 229	. . . 4 ⊢ (¬ 𝐴 ∈ 𝐵 → (𝐵 ∩ {𝐴}) = ∅)
7	6	uneq2d 4098	. . 3 ⊢ (¬ 𝐴 ∈ 𝐵 → ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = ((𝐵 ∩ 𝐴) ∪ ∅))
8		un0 4322	. . 3 ⊢ ((𝐵 ∩ 𝐴) ∪ ∅) = (𝐵 ∩ 𝐴)
9	7, 8	eqtrdi 2790	. 2 ⊢ (¬ 𝐴 ∈ 𝐵 → ((𝐵 ∩ 𝐴) ∪ (𝐵 ∩ {𝐴})) = (𝐵 ∩ 𝐴))
10	4, 9	eqtrid 2786	1 ⊢ (¬ 𝐴 ∈ 𝐵 → (𝐵 ∩ suc 𝐴) = (𝐵 ∩ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1547   ∈ wcel 2119   ∪ cun 3881   ∩ cin 3882  ∅c0 4261  {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-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-ral 3054  df-rab 3392  df-v 3433  df-dif 3886  df-un 3888  df-in 3890  df-nul 4262  df-sn 4556  df-suc 6316

This theorem is referenced by:  ackbij1lem15  10146  ackbij1lem16  10147