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Proof step 10 from proof div-ls-mone (previous step)

i0 < 0, 1 + i0 * j < j, i0 < j, 0 < j, i0 + 1 < j ⊦ i0 < ~1 ↔ i0 * j + 2 * j ≤ 0

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%mone-08: ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%minus-one-02: ⊦ ~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%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-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult%numls-10: ⊦ 0 < i → ((* i) * j < (* i) * j0 ↔ j < j0); from specification int-mult%irref: ⊦ ¬ i < i; from specification int-basic1