Metamath Proof Explorer


Theorem ackbij1lem1

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

Ref Expression
Assertion ackbij1lem1 ⊢ ¬ A ∈ B → B ∩ suc ⁡ 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 2 3 eqtri ⊢ B ∩ suc ⁡ A = B ∩ A ∪ B ∩ A
5 disjsn ⊢ B ∩ A = ∅ ↔ ¬ A ∈ B
6 5 biimpri ⊢ ¬ A ∈ B → B ∩ A = ∅
7 6 uneq2d ⊢ ¬ A ∈ B → B ∩ A ∪ B ∩ A = B ∩ A ∪ ∅
8 un0 ⊢ B ∩ A ∪ ∅ = B ∩ A
9 7 8 eqtrdi ⊢ ¬ A ∈ B → B ∩ A ∪ B ∩ A = B ∩ A
10 4 9 eqtrid ⊢ ¬ A ∈ B → B ∩ suc ⁡ A = B ∩ A