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Proof step 9 from proof div-ls-zero (previous step)

i0 * j + j0 < 0, j0 < j, i0 < j, 0 < j, abs(j0) < j, j0 ≤ 0, i0 ≤ 0 ⊦ i0 < 0 ↔ ¬ 0 < j * (i0 + 1) + j0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [j0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ¬ j < 0 → abs(j) = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ 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% ⊦ 0 < i → (0 < i * j ↔ 0 < j); (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ 0 < i ∧ ¬ j < 0 → 0 < i + j; (s) Used terms: j0