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Proof step 9 from proof div-eq-zero-01 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 10:
i < j, i / j = 0, i < 0, abs(j) + i ≤ 0 ⊦ 0 = j

Rule argument

fma: i < j

Additional information

Used simplifier rules: div-same: i ≠ 0 ⊦ i / i = 1;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%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-le: ⊦ i - j ≤ i0 ↔ i ≤ j + i0; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 Used formulas: i < j