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Proof step 13 from proof div-zero-02 (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 4

Additional information

Used simplifier rules: ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 ≤ i → abs(i) = i; (s)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i + 0 = i; (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 → ¬ j < i; (s)%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% ⊦ i + j < i + i0 ↔ j < i0; (s)% ⊦ i < abs(i) ↔ i < 0; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i + ~ j = i - j; (s)% ⊦ i < j - i0 ↔ i + i0 < j; (s)% ⊦ i ≤ j → ¬ j < i; (s)% ⊦ i < i + j ↔ 0 < j; (s)