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Proof step 2 from proof mod-twoodd-02 (previous step)

j % (2 * i) = 1, 0 < i, 0 ≤ j ⊦ j % 2 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
j % (2 * i) = 1, 0 < i, 0 ≤ j ⊦ j % 2 = 1

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs