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Proof step 4 from proof auxexdivmod-01 (previous step)

i * j ≤ i, 0 < j, 0 ≤ i ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: le-same seq: ⊦ 0 < i → (i * j ≤ i ↔ j ≤ 1) subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-le: ⊦ 0 < i → (i * j ≤ 0 ↔ j ≤ 0);%one: ⊦ 1 * i = i;%one-01: ⊦ i = 1 → i * j = j;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = 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%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%ler: ⊦ i ≤ i; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs Used terms: j, i