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Proof step 23 from proof int2bytes-last-01 (previous step)

~128 ≤ i % 256, i % 256 < 128, i < 0, i ≤ ~256, i < ~256 ⊦ i % 256 = 0, i2b(i % 256) = i2b(i)

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] → [i, 256, i0, j] rotate: T spec: int-div inst: name: div-elim-negpos

Additional information

Used simplifier rules: ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i2b(i * 256 + j) = i2b(j);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;%⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0); (elim) Used terms: i, 256, i0, j