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Proof step 10 from proof div-eq-zero-01 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 11:
i / j = 0, i < 0, i < j ⊦ 0 = j, 0 < abs(j) + i

Rule argument

fma: i < j

Additional information

Used simplifier rules: div-same: i ≠ 0 ⊦ i / i = 1;% ⊦ i < j ↔ ¬ j ≤ i; (ss)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i + ~ j = i - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i - j ≤ i0 ↔ i ≤ j + i0; (s)% ⊦ i + 0 = i; (s)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s) Used formulas: i < j