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Proof step 9 from proof firstn-twice (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
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.head + firstn(n, x.tail);% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ x + y = x + z ↔ y = z; (s)% ⊦ a + x = a ' + x; (s)% ⊦ (a ' + x).tail = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).head = a; (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)