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Proof step 7 from proof bib-ls-same-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ abs(i) < j → i % j = i
Proof step 9:
¬ abs(i) < 256, i % 256 + 256 < i, i % 256 < ~128, i < 0 ⊦ i % 256 = 0

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