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Proof step 5 from proof int2bits-rec-neg-03 (previous step)

j < ~1, j < 0 ⊦ j % 2 = 0, j = ~1, j / 2 = (j - 1) / 2 + 1

User applied rule insert elim lemma

And closed the goal!

Rule argument

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, j0] rotate: T spec: int-div inst: name: div-elim-negpos

Additional information

Used simplifier rules: ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i < 0 → abs(i) = ~ i;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ i < 0 → (~2 < i ↔ i = ~1);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i - i1 - j = i - i1 + j;% ⊦ i * j - i = i * (j - 1);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;%j ≠ 0 ⊦ (i * j) / j = i;% ⊦ i + (j - i) = j;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ 2 * i = 0 ↔ i = 0;%⊦ 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, j0