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Proof step 2 from proof letwozero (previous step)

i < 0 ⊦ 0 ≤ i / 2 ↔ i = ~1

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: div-ngr-zero: ⊦ i < 0 ∧ 0 < j → (0 ≤ i / j ↔ abs(i) < j);%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%minus-one: ⊦ i < 0 → (~2 < i ↔ i = ~1); from specification int-basic1%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2