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Proof step 32 from proof rightshift-8-01 (previous step)

abs(j) = 1, i→n(j) = 1, i < ~128, ~256 < i, ~256 ≤ i, i < 0, ∀ i, j0. i→n(j0) < 1 → i→n(j0) < # int2bytes(i) → i >> 8 * abs(j0) = bytes2int(firstn(# int2bytes(i) - i→n(j0), int2bytes(i))) ⊦ j = 0, i % 256 = 0, i / 256 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 33:
abs(i) < abs(j) ⊦ i / j = 0

Rule argument

seq: abs(i) < abs(j) ⊦ i / j = 0 subst: [j, i] → [256, i] rotate: F spec: int-div inst: name: div-zero-02

Additional information

Used simplifier rules: ⊦ i < 0 → abs(i) = ~ i;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: 256, i