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Proof step 12 from proof nodups-butlastn (previous step)

n ≤ # x0, n ≤ # x0 + 1, dups(a ' + butlastn(n, x0)), ∀ n0. ¬ n0 ≤ # x0 ∨ ¬ dups(butlastn(n0, x0)) ⊦ a ∈ x0, dups(x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
n ≤ # x0, n ≤ # x0 + 1, ∀ n0. ¬ n0 ≤ # x0 ∨ ¬ dups(butlastn(n0, x0)), a ∈ butlastn(n, x0) ∨ dups(butlastn(n, x0)) ⊦ a ∈ x0, dups(x0)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup