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Proof step 3 from proof maxScale-not-negt (previous step)

maxScale(i) < 0, ∀ i0. i→n(i0) < i→n(i) → true → ¬ maxScale(i0) < 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, maxScale(i / 10) + 1 < 0, ∀ i0. i→n(i0) < i→n(i) → true → ¬ maxScale(i0) < 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-zero: ⊦ maxScale(0) = 0;%maxScale-base: ⊦ i % 10 ≠ 0 ∧ i ≠ 0 → maxScale(i) = 0; Used terms: i