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Proof step 3 from proof divmod2-2pot-neg (previous step)

0 < i, 1 < i, j < 0 ⊦ ((j % 2 ^ i - 1) / 2) % 2 = ((j - 1) / 2) % 2

User applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
j = i0 * 2 ^ i + j0, j < 0, 1 < i, 0 < i, abs(j0) < 2 ^ i, 0 < 2 ^ i, j0 ≤ 0, i0 ≤ 0 ⊦ ((j0 - 1) / 2) % 2 = ((j - 1) / 2) % 2

Rule argument

spec: int-div name: div-elim-negpos seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0) subst: [i, j, i0, j0] → [j, 2 ^ i, i0, j0]

Additional information

Used simplifier rules: pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s)%⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0); (elim) Used terms: j, 2 ^ i, i0, j0