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Proof step 1 from proof lastN-rest (conclusion)

⊦ n < # x → lastn(n, x.rest) = lastn(n, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n < # x ⊦ restn(# x.rest - n, x.rest) = restn(# x - n, x), x = []

Rule argument

Additional information

Used simplifier rules: lastn-restn-01: n ≤ # x ⊦ lastn(n, x) = restn(# x - n, x);%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lr-05: ⊦ ¬ # x.rest + 1 < # x; from specification list-data%succ: ⊦ m +1 = m + 1; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%l: ⊦ # x = 0 ↔ x = []; from specification list-data%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%trans: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat