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Proof step 4 from proof firstn-twice (previous step)

n ≤ #(a + x0), ∀ n. n ≤ # x0 → firstn(n, firstn(n, x0)) = firstn(n, x0) ⊦ firstn(n, firstn(n, a + x0)) = firstn(n, a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n ≤ 1 + # x0, ∀ n. n ≤ # x0 → firstn(n, firstn(n, x0)) = firstn(n, x0) ⊦ firstn(n, firstn(n, a ' + x0)) = firstn(n, a ' + x0)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)