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Proof step 2 from proof rem-inter-empty3 (previous step)

(s \ s0) ∩ s1 = ∅, (s2 \ s) ∩ s1 = ∅ ⊦ (s2 \ s0) ∩ s1 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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