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Proof step 4 from proof nodups-butlastn (previous step)

dups(butlastn(n, a + x0)), n ≤ #(a + x0), ∀ n. dups(butlastn(n, x0)) ∧ n ≤ # x0 → dups(x0) ⊦ dups(a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
dups(butlastn(n, a ' + x0)), n ≤ 1 + # x0, ∀ n0. ¬ dups(butlastn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x; (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)