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Proof step 21 from proof div-add1-negpos-neg-01 (previous step)

i1 * j + i2 * j + j0 = i0 * j + j1, i1 * j < j1, i1 < j + j1, i1 + 1 < j + j1, i1 * j + i2 * j < j + j1, (i1 * j + i2 * j) + 1 < j + j1, i0 < j + j1, j1 < 0, j1 + 1 < j0, 0 < j + j1, i1 * j + i2 * j < j, i1 * j + i2 * j < j0, 2 + i1 < j, i1 < j0, 0 ≤ i2, 0 < j, 1 < j, j + i1 * j ≤ 0, i1 < 0, i1 < j, j0 < j, 0 < j0, i1 * j + i2 * j < ~1, 0 ≤ i2 * j + j0, j + i1 * j < j0, i1 * j + i2 * j < 0, 2 + i1 * j + i2 * j < j, j1 < j, j1 < j0, i0 ≤ 0, i0 < j, i0 < j0, i0 * j + j1 < 0, i0 * j + j1 < j, i0 * j + j1 < j0, 2 + i0 * j + j1 < j, i0 + 1 < j, 2 * j + j1 < j, 2 * j + j1 ≤ j0, j1 + j ≤ j0, j1 ≤ j0, (i0 * j + j1) + 1 < j0, i1 * j < j0, i1 + 1 < j0, (i1 * j + i2 * j) + 1 < j0, 2 + j1 < j ⊦ 2 * j + ~ j ≤ j0, j0 = 0, j = 1, i0 = i1 + i2 + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: ple-03 seq: i + j0 ≤ j, i0 ≤ j0 ⊦ i + i0 ≤ j subst: [i, j0, j, i0] → [2 * j, j1, j0, ~ j] flags: rewrite

Additional information

Used simplifier rules: neg-15: ⊦ i + ~ j = 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%sub-le: ⊦ i - j ≤ i0 ↔ i ≤ j + i0; from specification int-abs%numle-05: ⊦ (* i2) * i ≤ i + j0 ↔ (i2 - 1) * i ≤ j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%negle-13: ⊦ ~ i ≤ j ↔ 0 ≤ i + j; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: 2 * j, j1, j0, ~ j