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Proof step 1 from proof split-first (conclusion)

n ≤ # x ⊦ ∃ y, z. x = y + z ∧ # y = n

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ n. n ≤ # x0 → (∃ y, z. x0 = y + z ∧ # y = n), n ≤ #(a + x0) ⊦ ∃ y, z. a + x0 = y + z ∧ # y = n

Rule argument

indvar: x

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%list freely generated by [], +;;%size-01: ⊦ # [] = 0; from specification list-data%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%l: ⊦ # x = 0 ↔ x = []; from specification list-data