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Proof step 3 from proof mod-twoodd-01 (previous step)

∃ j. 0 ≤ j ∧ i = 2 * j + 1, i % 2 = 1, 0 ≤ i, 0 < j ⊦ i % (2 * j) = (i / 2) % j * 2 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 ≤ j0, 0 < j ⊦ (2 * j0 + 1) % (2 * j) = j0 % j * 2 + 1

Rule argument

Additional information

Used simplifier rules: mod-two-04: ⊦ 0 ≤ i → (i * 2 + 1) % 2 = 1;%div-two: ⊦ 0 ≤ i → (2 * i + 1) / 2 = i;%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%zero-ge-add-01: ⊦ 0 ≤ j ∧ 0 ≤ j0 ∧ 0 ≤ i0 → 0 ≤ i0 * j + j0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult