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Metamath Proof Explorer

Theorem uncom

Description: Commutative law for union of classes. Exercise 6 of TakeutiZaring p. 17. (Contributed by NM, 25-Jun-1998) (Proof shortened by Andrew Salmon, 26-Jun-2011)

		Ref	Expression
	Assertion	uncom	⊢ ( 𝐴 ∪ 𝐵 ) = ( 𝐵 ∪ 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		orcom	⊢ ( ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐴 ) )
2		elun	⊢ ( 𝑥 ∈ ( 𝐵 ∪ 𝐴 ) ↔ ( 𝑥 ∈ 𝐵 ∨ 𝑥 ∈ 𝐴 ) )
3	1 2	bitr4i	⊢ ( ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) ↔ 𝑥 ∈ ( 𝐵 ∪ 𝐴 ) )
4	3	uneqri	⊢ ( 𝐴 ∪ 𝐵 ) = ( 𝐵 ∪ 𝐴 )