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Proof step 5 from proof rem-inter-empty3 (previous step)

(s2 \ s) ∩ s1 = ∅, (s \ s0) ∩ s1 = ∅ ⊦ #((s2 \ s0) ∩ s1) = 0, ∀ a. a ∈ (s2 \ s0) ∩ s1 ↔ a ∈ ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
(s2 \ s) ∩ s1 = ∅, (s \ s0) ∩ s1 = ∅, a ∈ s2, a ∈ s1 ⊦ #((s2 \ s0) ∩ s1) = 0, # s2 = 0, # s1 = 0, s1 = ∅, s2 = ∅, a ∈ s0

Rule argument

Additional information

Used simplifier rules: Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic% ⊦ ¬ a ∈ ∅; (s)