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Metamath Proof Explorer

Theorem unissd

Description: Subclass relationship for subclass union. Deduction form of uniss . (Contributed by David Moews, 1-May-2017)

Ref Expression
Hypothesis unissd.1 ⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
Assertion unissd ( 𝜑 → ∪ 𝐴 ⊆ ∪ 𝐵 )

Proof

Step Hyp Ref Expression
1 unissd.1 ⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
2 uniss ⊢ ( 𝐴 ⊆ 𝐵 → ∪ 𝐴 ⊆ ∪ 𝐵 )
3 1 2 syl ⊢ ( 𝜑 → ∪ 𝐴 ⊆ ∪ 𝐵 )