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Proof step 1 from proof firstN-01 (conclusion)

⊦ n < # x ∧ list2set(x) ∩ s = ∅ ∧ list2set(x) ∩ s0 = ∅ → list2set(firstn(n, x)) ∩ (s ∪ s0) = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
list2set(x) ∩ s0 = ∅, list2set(x) ∩ s = ∅, n < # x ⊦ #(list2set(firstn(n, x)) ∩ (s ∪ s0)) = 0, list2set(firstn(n, x)) ∩ (s ∪ s0) = ∅, x = []

Rule argument

Additional information

Used simplifier rules: ⊦ # s ≠ 0 → s ≠ ∅; (forward)% ⊦ # x = 0 ↔ x = [];% ⊦ s ≠ ∅ → # s ≠ 0; (forward)% ⊦ m < n → n ≠ 0; (forward)