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Proof step 11 from proof getint-neg-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ abs(i) < j → i % j = i

Rule argument

spec: int-div inst: name: mod-zero-05 seq: ⊦ abs(i) < j → i % j = i subst: [j, i] → [10, i] flags: rewrite

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%numlelsdel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* 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 Used terms: 10, i