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Proof step 5 from proof div-mult-two-nls-01 (previous step)

j0 < j, 1 + 2 * i0 < (2 * (i0 * j + j0)) / j, 0 < j, 0 ≤ j0, 0 ≤ i0 ⊦

User applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)
Proof step 7:
2 * (i0 * j + j0) = i1 * j + j1 ∧ 0 ≤ j1 ∧ j1 < j ∧ 0 ≤ i1 ∧ 0 ≤ 2 * (i0 * j + j0), j0 < j, 1 + 2 * i0 < i1, 0 < j, 0 ≤ j0, 0 ≤ i0 ⊦

Rule argument

spec: int-div inst: name: div-elim-pospos seq: ⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0) subst: [i, j, i0, j0] → [2 * (i0 * j + j0), j, i, j1] flags: current, []

Additional information

Used simplifier rules: zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%zero-ge-add: ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); from specification int-mult%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: 2 * (i0 * j + j0), j, i, j1