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Proof step 16 from proof maxScale-pot10-01 (previous step)

0 < j, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ j0 → maxScale(i * 10 ^ j0) = j0 ∨ i % 10 = 0 ⊦ i % 10 = 0, maxScale(i * 10 ^ (j - 1)) + 1 = j, i = 0, j = 0

User applied rule apply induction

And closed the goal!

Rule argument

subst: [j0, i] → [j - 1, i]

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat Used terms: j - 1, i