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Proof step 6 from proof restN-dups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ # x0 + 1, dups(restn(n - 1, x0)), ∀ n0. ¬ dups(restn(n0, x0)) ∨ ¬ n0 ≤ # x0 ⊦ n = 0, a ∈ x0, dups(x0)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%pred-03: ⊦ n -1 = n - 1; from specification nat%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1