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Proof step 1 from proof intempty-02 (conclusion)

⊦ list2set(x) ∩ s = ∅ ∧ a ∈ s → ¬ a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
list2set(x) ∩ s = ∅, a ∈ s, a ∈ x ⊦ s = ∅, # s = 0

Rule argument

Additional information

Used simplifier rules: ⊦ s ≠ ∅ → # s ≠ 0; (forward)% ⊦ a ∈ s → s ≠ ∅; (forward)% ⊦ s ∩ s0 = s0 ∩ s;% ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;