Metamath Proof Explorer


Theorem unass

Description: Associative law for union of classes. Exercise 8 of TakeutiZaring p. 17. (Contributed by NM, 3-May-1994) (Proof shortened by Andrew Salmon, 26-Jun-2011)

Ref Expression
Assertion unass ⊢ A ∪ B ∪ C = A ∪ B ∪ C

Proof

Step Hyp Ref Expression
1 elun ⊢ x ∈ A ∪ B ∪ C ↔ x ∈ A ∨ x ∈ B ∪ C
2 elun ⊢ x ∈ B ∪ C ↔ x ∈ B ∨ x ∈ C
3 2 orbi2i ⊢ x ∈ A ∨ x ∈ B ∪ C ↔ x ∈ A ∨ x ∈ B ∨ x ∈ C
4 elun ⊢ x ∈ A ∪ B ↔ x ∈ A ∨ x ∈ B
5 4 orbi1i ⊢ x ∈ A ∪ B ∨ x ∈ C ↔ x ∈ A ∨ x ∈ B ∨ x ∈ C
6 orass ⊢ x ∈ A ∨ x ∈ B ∨ x ∈ C ↔ x ∈ A ∨ x ∈ B ∨ x ∈ C
7 5 6 bitr2i ⊢ x ∈ A ∨ x ∈ B ∨ x ∈ C ↔ x ∈ A ∪ B ∨ x ∈ C
8 1 3 7 3bitrri ⊢ x ∈ A ∪ B ∨ x ∈ C ↔ x ∈ A ∪ B ∪ C
9 8 uneqri ⊢ A ∪ B ∪ C = A ∪ B ∪ C