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Proof step 6 from proof maxScale-not-negt (previous step)

i % 10 = 0, maxScale(i / 10) < ~1, maxScale(i / 10) < 0, ∀ i0. i→n(i0) < i→n(i) → true → ¬ maxScale(i0) < 0 ⊦ i = 0

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i0] → [i / 10]

Additional information

Used simplifier rules: labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abs: j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j); from specification int-div%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%abs-div10-le: ⊦ abs(i) ≤ abs(i) / 10 ↔ i = 0; from specification int-div Used terms: i / 10