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Proof step 3 from proof div-ls-mone (previous step)

0 < j ⊦ i / j < ~1 ↔ i + 2 * j ≤ 0

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: i < 0

Additional information

Used simplifier rules: div-zero-nls: ⊦ 0 ≤ i ∧ 0 < j → ¬ i / j < 0;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%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%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%lep-01: ⊦ i + j ≤ i ↔ j ≤ 0; from specification int-abs%zero-le: ⊦ 0 < i → (i * j ≤ 0 ↔ j ≤ 0); from specification int-mult%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); 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 formulas: i < 0