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Proof step 1 from proof zero-ge-02 (conclusion)

⊦ 0 ≤ i → (0 ≤ i * j ↔ ¬ ¬(i = 0 ∨ 0 ≤ j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i ⊦ 0 ≤ i * j ↔ i = 0 ∨ 0 ≤ j

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs