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Proof step 3 from proof shiftright-neg-div-02 (previous step)

i < ~(2 ^ (abs(j) - 1)) ∧ j ≠ 0 ⊦ i % 2 ^ abs(j) ≠ 0 ∧ i < ~(2 ^ (abs(j) - 1)) ∧ j ≠ 0 → (i - (2 ^ abs(j) - 1)) / 2 ^ abs(j) = i / 2 ^ abs(j) - 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i % j ≠ 0 ∧ j0 = j - 1 ∧ i < 0 ∧ 0 < j → (i - j0) / j = i / j - 1

Rule argument

seq: ⊦ i % j ≠ 0 ∧ j0 = j - 1 ∧ i < 0 ∧ 0 < j → (i - j0) / j = i / j - 1 subst: [j, j0, i] → [2 ^ abs(j), 2 ^ abs(j) - 1, i] rotate: F spec: int-div inst: name: div-minus-onediff

Additional information

Used simplifier rules: ⊦ i - (j - j0) = i + j0 - j;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i < ~ j ↔ i + j < 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 0 ≤ abs(i);% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i + j < j ↔ i < 0;% ⊦ ¬ 2 ^ i < 0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: 2 ^ abs(j), 2 ^ abs(j) - 1, i