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Proof step 6 from proof nodups-butlastn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: butlastN-rec: ⊦ butlastn(n + 1, x) = butlastn(n, x.butlast);%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%les: ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat