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Proof step 3 from proof maxScale-zero-01 (previous step)

(i / 10 ^ maxScale(i)) % 10 = 0, ∀ i0. i→n(i0) < i→n(i) → (i0 / 10 ^ maxScale(i0)) % 10 = 0 → i0 = 0 ⊦ i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i % 10 = 0 ∧ i ≠ 0 → maxScale(i) = maxScale(i / 10) + 1
Proof step 5:
i % 10 = 0 ∧ i ≠ 0, (i / 10 ^ (maxScale(i / 10) + 1)) % 10 = 0, ∀ i0. i→n(i0) < i→n(i) → (i0 / 10 ^ maxScale(i0)) % 10 = 0 → i0 = 0 ⊦ i = 0

Rule argument

spec: inst: name: maxScale-rec seq: ⊦ i % 10 = 0 ∧ i ≠ 0 → maxScale(i) = maxScale(i / 10) + 1 subst: [i] → [i] flags: rewrite, current

Additional information

Used simplifier rules: maxScale-base: ⊦ i % 10 ≠ 0 ∧ i ≠ 0 → maxScale(i) = 0;%one: ⊦ i / 1 = i; from specification int-div Used terms: i