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Proof step 3 from proof nodups-butlastn (previous step)

dups(butlastn(n, x)), n ≤ # x ⊦ dups(x)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
dups(butlastn(n, a + x0)), n ≤ #(a + x0), ∀ n. dups(butlastn(n, x0)) ∧ n ≤ # x0 → dups(x0) ⊦ dups(a + x0)

Rule argument

term: x

Additional information

Used simplifier rules: butlastN-base: ⊦ butlastn(0, x) = x;%list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ ¬ dups([]); (s)