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Proof step 4 from proof firstN-01 (previous step)

list2set(x) ∩ s0 = ∅, list2set(x) ∩ s = ∅, n < # x ⊦ #(list2set(firstn(n, x)) ∩ (s ∪ s0)) = 0, ∀ a. a ∈ list2set(firstn(n, x)) ∩ (s ∪ s0) ↔ a ∈ ∅, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
list2set(x) ∩ s = ∅, list2set(x) ∩ s0 = ∅, a ∈ x, a ∈ firstn(n, x), a ∈ s ∪ s0, n < # x ⊦ s ∪ s0 = ∅, #(list2set(firstn(n, x)) ∩ (s ∪ s0)) = 0, x = [], #(s ∪ s0) = 0

Rule argument

Additional information

Used simplifier rules: in: ⊦ a ∈ list2set(x) ↔ a ∈ x;% ⊦ s ≠ ∅ → # s ≠ 0; (forward)% ⊦ a ∈ s → s ≠ ∅; (forward)% ⊦ a ∈ firstn(n, x) ∧ n < # x → a ∈ x; (forward)% ⊦ s ∩ s0 = s0 ∩ s;% ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;% ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;% ⊦ ¬ a ∈ ∅;