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Proof step 63 from proof div-div-rec (previous step)

2 ^ i * ((2 * i1 + j1) + 1) + j0 < 0, abs(j1) < 2, 2 * i1 + j1 < 0, j1 < i, i1 < i, (2 * i1 + j1) + 1 < i, j0 < i, abs(j0) < 2 ^ i, 0 < i, 2 ^ i * (j1 + 2 * (i1 + 1)) + j0 < i, 2 ^ i * j1 + j0 < i, j0 ≤ 0, 0 ≤ i, j1 ≤ 0, i1 ≤ 0, 2 ^ i * (j1 + 2 * (i1 + 1)) + j0 ≤ 0, 2 ^ i * j1 + j0 ≤ 0, 2 ^ i * 2 ≤ abs(2 ^ i * j1 + j0) ⊦ i1 = (2 ^ i * (j1 + 2 * i1) + j0) / (2 ^ i * 2), abs(j1) ≠ 1 ∧ 0 ≤ abs(j1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 64:
abs(j1) = 1, 2 ^ i * ((2 * i1 + j1) + 1) + j0 < 0, 2 * i1 + j1 < 0, j1 < i, i1 < i, (2 * i1 + j1) + 1 < i, j0 < i, abs(j0) < 2 ^ i, 0 < i, 2 ^ i * (j1 + 2 * (i1 + 1)) + j0 < i, 2 ^ i * j1 + j0 < i, j0 ≤ 0, 0 ≤ i, j1 ≤ 0, i1 ≤ 0, 2 ^ i * (j1 + 2 * (i1 + 1)) + j0 ≤ 0, 2 ^ i * j1 + j0 ≤ 0, 2 ^ i * 2 ≤ abs(2 ^ i * j1 + j0) ⊦ i1 = (2 ^ i * (j1 + 2 * i1) + j0) / (2 ^ i * 2)

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1% ⊦ 0 ≤ abs(i); (s)