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Proof step 5 from proof sublist-at (previous step)

n0 < n, m + n ≤ # x, n ≤ # restn(m, x) ⊦ restn(m, x)[n0] = x[m + n0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 < n, m + n ≤ # x, n + m ≤ # x ⊦ restn(m, x)[n0] = x[m + n0]

Rule argument

Additional information

Used simplifier rules: rest-len: n ≤ # x ⊦ # restn(n, x) = # x - n;%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m ≤ n ↔ ¬ n < m; (ss)