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Proof step 4 from proof div-zero-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
0 < j, i < 0, abs(i) < j ⊦ i / j = 0
Proof step 6:
abs(i) < abs(j) ⊦ i / j = 0, 0 < j ∧ 0 ≤ i ∧ i < j, 0 < j ∧ i < 0 ∧ abs(i) < j

Rule argument

spec: <toplevel> name: div-zero-01 seq: 0 < j, i < 0, abs(i) < j ⊦ i / j = 0 subst: [j, i] → [j, i] flags: <all paths>

Additional information

Used terms: j, i