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Proof step 2 from proof subseteq-inter-empty-empty (previous step)

s1 ∩ s2 = ∅, s0 ⊆ s1 ⊦ s0 ∩ s2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
s1 ∩ s2 = ∅, s0 ⊆ s1 ⊦ #(s0 ∩ s2) = 0, s0 ∩ s2 = ∅

Rule argument

Additional information

Used simplifier rules: size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic