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Proof step 1 from proof restn-less (conclusion)

⊦ n < # x → restn(n, x) = x[n] + restn(n +1, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: restN-rec: ⊦ restn(n + 1, x) = restn(n, x.rest);%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ a + x = a ' + x; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2