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Proof step 3 from proof is-div (previous step)

0 < j, i < (i0 + 1) * j, 0 ≤ i0, 0 ≤ i, i0 * j ≤ i ⊦ i0 = i / j

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: <toplevel> name: Div-def seq: ⊦ 0 ≤ i ∧ 0 < j → i / j * j ≤ i ∧ i < (i / j)+1 * j subst: [i, j] → [i, j]

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i +1 = i + 1; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ 0 < i → (0 < i * j ↔ 0 < j); (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); (s)% ⊦ i < j + 1 ↔ i ≤ j; (s)% ⊦ i ≤ i; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < j → ¬ j < i; (s) Used terms: i, j