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Proof step 3 from proof zero (previous step)

j < 0 ⊦ 0 / j = 0, j = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ 0 ≤ i ∧ 0 < j → i / j * j ≤ i ∧ i < (i / j)+1 * j

Rule argument

spec: int-div inst: name: Div-def seq: ⊦ 0 ≤ i ∧ 0 < j → i / j * j ≤ i ∧ i < (i / j)+1 * j subst: [i, j] → [0, ~ j] flags: current, []

Additional information

Used simplifier rules: Div-right: ⊦ j ≠ 0 → i / ~ j = ~(i / j);%div-zero-ngr-01: ⊦ 0 ≤ i ∧ j < 0 → ¬ 0 < i / j;%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%neg-02: ⊦ ~ ~ i = i; from specification int-abs%zero-le-01: ⊦ i < 0 → (i * j ≤ 0 ↔ 0 ≤ j); from specification int-mult%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs Used terms: 0, ~ j