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Proof step 3 from proof to-plus-from-fin (previous step)

n ≤ # x ⊦ (x to n) + (x from n) = x

User applied rule structural induction

And obtained the following new goals

Proof step 4:
n ≤ #(a + x0), ∀ n. n ≤ # x0 → (x0 to n) + (x0 from n) = x0 ⊦ (a + x0 to n) + (a + x0 from n) = a + x0

Rule argument

indvar: x

Additional information

Used simplifier rules: to-zero-l: ⊦ x to 0 = [];%from-zero-l: ⊦ x from 0 = x;%list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list