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Proof step 5 from proof bib-lower (previous step)

~256 < i, i < ~128, i < 0, ~128 ≤ i % 256, i < i % 256 ⊦ i % 256 = i + 256

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ abs(i) < j → i % j = i

Rule argument

seq: ⊦ abs(i) < j → i % j = i subst: [j, i] → [256, i] rotate: F spec: int-div inst: name: mod-zero-05

Additional information

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