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Metamath Proof Explorer

Theorem unss2

Description: Subclass law for union of classes. Exercise 7 of TakeutiZaring p. 18. (Contributed by NM, 14-Oct-1999)

Ref Expression
Assertion unss2 ( 𝐴 ⊆ 𝐵 → ( 𝐶 ∪ 𝐴 ) ⊆ ( 𝐶 ∪ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 unss1 ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝐴 ∪ 𝐶 ) ⊆ ( 𝐵 ∪ 𝐶 ) )
2 uncom ⊢ ( 𝐶 ∪ 𝐴 ) = ( 𝐴 ∪ 𝐶 )
3 uncom ⊢ ( 𝐶 ∪ 𝐵 ) = ( 𝐵 ∪ 𝐶 )
4 1 2 3 3sstr4g ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝐶 ∪ 𝐴 ) ⊆ ( 𝐶 ∪ 𝐵 ) )