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Proof step 2 from proof firstn-last-rec (previous step)

∀ x. n0 ≠ 0 ∧ n0 ≤ # x → firstn(n0, x) = firstn(n0 -1, x) + x[n0 -1] ⊦ n0 +1 ≠ 0 ∧ n0 +1 ≤ # x → firstn(n0 +1, x) = firstn(n0 +1 -1, x) + x[n0 +1 -1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 < # x, ∀ x0. n0 ≠ 0 ∧ n0 ≤ # x0 → firstn(n0, x0) = firstn(n0 - 1, x0) + x0[n0 - 1] ' ⊦ x.first ' + firstn(n0, x.rest) = firstn(n0, x) + x[n0] ', x = []

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%pred-03: ⊦ n -1 = n - 1; from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%Last: ⊦ x + a = x + a '; from specification list%pred: ⊦ (m + 1)-1 = m; from specification nat-basic2%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%succ: ⊦ m +1 = m + 1; from specification nat-basic2