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Proof step 2 from proof div-shift-two (previous step)

i0 = (i / 2) / j * 2, 0 ≤ j, 0 ≤ i ⊦ i / j = i0 + (i - j * i0 < j ⊃ 0;1), j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 < j, 0 ≤ i ⊦ i / j = i / (2 * j) * 2 + (i < j * (i / (2 * j) * 2 + 1) ⊃ 0;1), j = 0

Rule argument

Additional information

Used simplifier rules: div-div-pos: ⊦ 0 ≤ i ∧ 0 < j ∧ 0 < j0 → (i / j) / j0 = i / (j * j0);%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% ⊦ 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 * j + i = i * (j + 1); (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)