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Proof step 6 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 7:
⊦ i + j = j + i
Proof step 8:
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 = (((j0 - 1) + i0 * 2 ^ i) / 2) % 2

Rule argument

spec: int-basic2 inst: name: add-c seq: ⊦ i + j = j + i subst: [j, i] → [j0 - 1, i0 * 2 ^ i] flags: rewrite

Additional information

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