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Proof step 11 from proof firstN-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
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

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ ∅;