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Proof step 12 from proof firstN-01 (previous step)

list2set(x) ∩ s0 = ∅, n < # x, a ∈ firstn(n, x), a ∈ x, a ∈ s, ∀ a0. ¬ a0 ∈ list2set(x) ∩ s ⊦ s = ∅, #(s ∪ s0) = 0, x = [], #(list2set(firstn(n, x)) ∩ (s ∪ s0)) = 0, s ∪ s0 = ∅, # s = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: a

Additional information

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