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Proof step 5 from proof firstn-last-rec (previous step)

n0 = m + 1, n0 < # x, ∀ x0. n0 ≤ # x0 → firstn(n0, x0) = firstn(m, x0) + x0[m] ' ∨ n0 = 0 ⊦ x.head ' + firstn(n0, x.tail) = firstn(n0, x) + x[n0] ', n0 = 0, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m + 1 < # x, ∀ x0. m < # x0 → x0.head ' + firstn(m, x0.tail) = firstn(m, x0) + x0[m] ' ⊦ x.tail.head ' + firstn(m, x.tail.tail) = firstn(m, x.tail) + x.tail[m] ', x = []

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);%get-rec-01: ⊦ x ≠ [] → x[n + 1] = x.tail[n];% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ x + y = x + z ↔ y = z; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)