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Proof step 1 from proof diff-sub-size (conclusion)

s = s0 \ s1, s1 ⊆ s0 ⊦ # s = # s0 - # s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ a. a ∈ s1 → a ∈ s0 ⊦ #(s0 \ s1) = # s0 - # s1

Rule argument

Additional information

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