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Proof step 6 from proof firstn-twice (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ # x0 + 1, ∀ n. n ≤ # x0 → firstn(n, firstn(n, x0)) = firstn(n, x0) ⊦ n = 0, firstn(n, a ' + firstn(n - 1, x0)) = a ' + firstn(n - 1, x0)

Rule argument

Additional information

Used simplifier rules: pred-03: ⊦ n -1 = n - 1; from specification nat