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Proof step 4 from proof divmod2-2pot-neg (previous step)

0 ≤ i, i0 < i, i0 ≤ 0, j0 < i, j0 ≤ 0, i0 * 2 ^ i + j0 < 0, abs(j0) < 2 ^ i, 1 < i, 0 < i ⊦ ((j0 - 1) / 2) % 2 = ((i0 * 2 ^ i + j0 - 1) / 2) % 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ i + (j - i0) = i + j - i0
Proof step 6:
0 ≤ i, i0 < i, i0 ≤ 0, j0 < i, j0 ≤ 0, i0 * 2 ^ i + j0 < 0, abs(j0) < 2 ^ i, 1 < i, 0 < i ⊦ ((j0 - 1) / 2) % 2 = ((i0 * 2 ^ i + (j0 - 1)) / 2) % 2

Rule argument

spec: int-abs inst: name: sub-add-01 seq: ⊦ i + (j - i0) = i + j - i0 subst: [i0, j, i] → [1, j0, i0 * 2 ^ i] flags: rewrite

Additional information

Used terms: 1, j0, i0 * 2 ^ i