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Proof step 7 from proof restN-dups (previous step)

0 < n, dups(restn(n -1, (a ' + x0).tail)), n ≤ 1 + # x0, ∀ n0. ¬ dups(restn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ dups(x0), a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
dups(restn(n - 1, x0)), n ≤ 1 + # x0, ∀ n0. ¬ dups(restn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ n = 0, dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < m ↔ m ≠ 0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)% ⊦ n -1 = n - 1; (s)