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Proof step 10 from proof div-zero-02 (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 11:
0 < j, 0 ≤ i, i < j ⊦ i / j = 0
Proof step 12:
abs(i) < abs(j) ⊦ i / j = 0, 0 < j ∧ i < 0 ∧ ~ i < j, j < 0 ∧ 0 ≤ i ∧ i + j < 0, 0 < j ∧ 0 ≤ i ∧ i < j, 0 < ~ j ∧ 0 ≤ ~ i ∧ ~ i < ~ j

Rule argument

spec: <toplevel> name: div-zero seq: 0 < j, 0 ≤ i, i < j ⊦ i / j = 0 subst: [j, i] → [~ j, ~ i]

Additional information

Used simplifier rules: neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%neg: j ≠ 0 ⊦ i / ~ j = ~(i / j);% ⊦ ~ i = 0 ↔ i = 0; (s)% ⊦ ¬ abs(i) < 0; (s)% ⊦ ~ ~ i = i; (s) Used terms: ~ j, ~ i