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Proof step 1 from proof union-sizele (conclusion)

⊦ #(s ∪ s0) ≤ # s ↔ s0 ⊆ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ #(s ∪ s0) ≤ # s ↔ (∀ a. a ∈ s0 → a ∈ s)

Rule argument

Additional information

Used simplifier rules: Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); from specification set-basic