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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 ) )