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Proof step 22 from proof bib-ls-same-01 (previous step)

0 < i % 256, 0 ≤ i, 127 < i % 256 ⊦ i % 256 < 256 + i ↔ ¬ i < 128

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ abs(i) < j → i % j = i
Proof step 24:
¬ abs(i) < 256, 0 < i % 256, 0 ≤ i, 127 < i % 256 ⊦ i % 256 < 256 + i ↔ ¬ i < 128

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: ⊦ 0 < i → i1 < i1 + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + j ≤ i ↔ j ≤ 0;% ⊦ 0 < j → abs(j) = j;%j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0);%i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j; Used terms: 256, i