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Proof step 8 from proof firstn-twice (previous step)

n = m + 1, n ≤ # x0 + 1, ∀ n. n ≤ # x0 → firstn(n, firstn(n, x0)) = firstn(n, x0) ⊦ n = 0, firstn(n, a ' + firstn(m, x0)) = a ' + firstn(m, x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
m ≤ # x0, ∀ n. n ≤ # x0 → firstn(n, firstn(n, x0)) = firstn(n, x0) ⊦ firstn(m, firstn(m, x0)) = firstn(m, x0)

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%c: ⊦ a + x = a ' + x; from specification list%c-08: ⊦ (a ' + x).first = a; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2