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