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Proof step 2 from proof mod-twoodd-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i - j = i0 ↔ i = i0 + j; (s)