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Proof step 3 from proof blastN-firstN (previous step)

n ≤ # x ⊦ butlastn(n, x) = firstn(# x - n, x)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
n ≤ #(a + x0), ∀ n. n ≤ # x0 → butlastn(n, x0) = firstn(# x0 - n, x0) ⊦ butlastn(n, a + x0) = firstn(#(a + x0) - n, a + x0)

Rule argument

indvar: x

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ butlastn(0, x) = x; (s)