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Proof step 5 from proof firstN (previous step)

s ∩ list2set(y) = ∅, m < # y, ∀ x0, s0. ¬ # x0 ≤ m ∧ list2set(x0) ∩ s0 = ∅ → list2set(firstn(m, x0)) ∩ s0 = ∅ ⊦ (list2set(firstn(m, y)) ++ a) ∩ s = ∅, #((list2set(firstn(m, y)) ++ a) ∩ s) = 0, y = [], a ∈ s

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: y, s

Additional information

Used simplifier rules: ⊦ s ∩ s0 = s0 ∩ s;% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m ≤ n ∧ n < n0 → m < n0; (forward)% ⊦ ¬ n < n;% ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;% ⊦ s ∩ s0 = ∅ ∧ ¬ a ∈ s → s ∩ (s0 ++ a) = ∅;% ⊦ # ∅ = 0; Used terms: y, s