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Proof step 5 from proof restn-empty (previous step)

0 < n, restn(n -1, (a ' + x0).rest) = [], n ≤ # x0, ∀ n0. n0 < # x0 → restn(n0, x0) = [] → n0 = # x0 ⊦ n = # x0 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
restn(n - 1, x0) = [], n ≤ # x0, ∀ n0. n0 < # x0 → restn(n0, x0) = [] → n0 = # x0 ⊦ n = # x0 + 1, n = 0

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%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