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Proof step 2 from proof firstn-prefix (previous step)

n ≤ # x ⊦ firstn(n, x) ⊑ x

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ n. n ≤ # x0 → firstn(n, x0) ⊑ x0, n ≤ #(a + x0) ⊦ firstn(n, a + x0) ⊑ a + x0

Rule argument

indvar: x

Additional information

Used simplifier rules: firstN-empty: ⊦ firstn(n, x) = [] ↔ n = 0;%list freely generated by [], +;;%size-01: ⊦ # [] = 0; from specification list-data%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%prefix-empty: ⊦ x ⊑ [] ↔ x = []; from specification list-last