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Proof step 41 from proof equal-minus-one (previous step)

1 / i ^ ~ j = ~1, j < 0, i ^ ~ j < 0 ⊦ i ^ ~ j = ~1, i = 0, i = ~1, j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 42:
abs(i) < abs(j) ⊦ i / j = 0

Rule argument

spec: int-div name: div-zero-02 seq: abs(i) < abs(j) ⊦ i / j = 0 subst: [j, i] → [i ^ ~ j, 1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i < ~ j ↔ i + j < 0; (s)% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: i ^ ~ j, 1