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Proof step 4 from proof mod-twoodd (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < j, 0 ≤ i ⊦ i = (2 * i + 1) / (2 * j) * j + i % j, j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 2 * i = 0 ↔ i = 0; (s)% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); (s)% ⊦ i + j = i + i0 ↔ j = i0; (s)%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% ⊦ i * j + i * j0 = i * (j + j0); (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)