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Proof step 5 from proof le-abs-numerator (previous step)

0 ≤ i ∧ 0 < j → i / j * j ≤ i ⊦ j = 0, 0 ≤ abs(i) + i / j ∧ i / j ≤ abs(i)

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: ¬ i < 0 ∧ ¬ j < 0

Additional information

Used simplifier rules: div-zero-nls: ⊦ 0 ≤ i ∧ 0 < j → ¬ i / j < 0;%div-ls-same: ⊦ 0 < j → (i < i / j ↔ i < 0 ∧ 1 < j);%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs Used formulas: ¬ i < 0 ∧ ¬ j < 0