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Proof step 2 from proof abs-intro (previous step)

0 ≤ i ∧ 0 < j ∨ i < 0 ∧ j < 0 ⊦ i / j = abs(i) / abs(j)

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%neg: j ≠ 0 ⊦ i / ~ j = ~(i / j);%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg-10: ⊦ ~ i = 0 ↔ i = 0; from specification int-abs%neg-02: ⊦ ~ ~ i = i; from specification int-abs%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs