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Proof step 2 from proof lastN-rest (previous step)

n < # x ⊦ lastn(n, x.tail) = lastn(n, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # x ⊦ restn(# x.tail - n, x.tail) = restn(# x - n, x), x = []

Rule argument

Additional information

Used simplifier rules: lastn-restn-01: n ≤ # x ⊦ lastn(n, x) = restn(# x - n, x);% ⊦ m < n → ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ # x.tail + 1 < # x; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ # x = 0 ↔ x = []; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)