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Proof step 28 from proof div-two-intro (previous step)

i = i0 * 2 + j1, j1 < 2, 0 < 2, 0 ≤ j1, 0 ≤ i0, 0 ≤ i ⊦ i / (j0 * 2) = i0 / j0, j0 = 0, j1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
0 ≤ i0 ⊦ (i0 * 2 + 1) / (j0 * 2) = i0 / j0, j0 = 0

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% ⊦ 0 < j → (j < 2 ↔ j = 1); (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (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 ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); (s)