Metamath Proof Explorer


Theorem ackbij1lem2

Description: Lemma for ackbij2 . (Contributed by Stefan O'Rear, 18-Nov-2014)

Ref Expression
Assertion ackbij1lem2 ⊢ A ∈ B → B ∩ suc ⁡ A = A ∪ B ∩ A

Proof

Step Hyp Ref Expression
1 df-suc ⊢ suc ⁡ A = A ∪ A
2 1 ineq2i ⊢ B ∩ suc ⁡ A = B ∩ A ∪ A
3 indi ⊢ B ∩ A ∪ A = B ∩ A ∪ B ∩ A
4 uncom ⊢ B ∩ A ∪ B ∩ A = B ∩ A ∪ B ∩ A
5 2 3 4 3eqtri ⊢ B ∩ suc ⁡ A = B ∩ A ∪ B ∩ A
6 snssi ⊢ A ∈ B → A ⊆ B
7 sseqin2 ⊢ A ⊆ B ↔ B ∩ A = A
8 6 7 sylib ⊢ A ∈ B → B ∩ A = A
9 8 uneq1d ⊢ A ∈ B → B ∩ A ∪ B ∩ A = A ∪ B ∩ A
10 5 9 eqtrid ⊢ A ∈ B → B ∩ suc ⁡ A = A ∪ B ∩ A