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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n ≤ # x0 + 1, dups(butlastn(n, a ' + x0)), ∀ n0. ¬ n0 ≤ # x0 ∨ ¬ dups(butlastn(n0, x0)) ⊦ a ∈ x0, dups(x0)

Rule argument

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list