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Proof step 21 from proof div-add1-posneg-pos-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: ⊦ i + ~ j = i - j; (s)% ⊦ i * j - i = i * (j - 1); (s)% ⊦ 1 * i = i; (s)%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% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ ~ i ≤ j ↔ 0 ≤ i + j; (s)% ⊦ i < j → ¬ j < i; (s) Used terms: 2 * j, j0, j1, ~ j