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Proof step 4 from proof firstn-last-rec (previous step)

n0 = m + 1 ∧ n0 ≠ 0, n0 < # x, ∀ x0. n0 ≠ 0 ∧ n0 ≤ # x0 → firstn(n0, x0) = firstn(m, x0) + x0[m] ' ⊦ x.first ' + firstn(n0, x.rest) = firstn(n0, x) + x[n0] ', x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m + 1 < # x, ∀ x0. m < # x0 → x0.first ' + firstn(m, x0.rest) = firstn(m, x0) + x0[m] ' ⊦ x = [], x.rest.first ' + firstn(m, x.rest.rest) = firstn(m, x.rest) + x.rest[m] '

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%get-rec-01: ⊦ x ≠ [] → x[n + 1] = x.rest[n];%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ a + x = a ' + x; from specification list