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Proof step 9 from proof getint-neg-02 (previous step)

i % 10 = 0, ∀ i0, j. i→n(i0) < i→n(i) → true → (mkDecimal(i0, j).int < 0 ↔ i0 < 0) ⊦ j = 0, mkDecimal(i / 10, abs(j) - 1).int < 0 ↔ i < 0

User applied rule apply induction

And obtained the following new goals

Proof step 10:
mkDecimal(i / 10, abs(j) - 1).int < 0 ↔ i / 10 < 0, i % 10 = 0, ∀ i0, j. i→n(i0) < i→n(i) → true → (mkDecimal(i0, j).int < 0 ↔ i0 < 0) ⊦ j = 0, mkDecimal(i / 10, abs(j) - 1).int < 0 ↔ i < 0

Rule argument

subst: [i0, j] → [i / 10, abs(j) - 1]

Additional information

Used simplifier rules: intvalue-zero: ⊦ mkDecimal(0, j).int = 0;%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, abs(j) - 1