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Proof step 63 from proof bib-pos-high (previous step)

i < i % 256, 0 ≤ i, 127 < i % 256, i ≤ 127, 0 < i % 256 ⊦ i % 256 = i + 256

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 64:
⊦ 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: ⊦ 0 ≤ i → abs(i) = i;% ⊦ i = i + j ↔ j = 0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ 0 < j → abs(j) = j;%j0 ≠ 0, j < j0 ⊦ i ≤ (* j) → i < (* j0);% ⊦ ¬ i < i; Used terms: 256, i