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Proof step 2 from proof bib-upper (previous step)

0 < i, i < 256, 127 < i ⊦ i = b2i(i2b(i)) + 256

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i ∧ 127 < i % 256 → b2i(i2b(i)) = i % 256 - 256

Rule argument

name: bib-pos-high rotate: F subst: [i] → [i]

Additional information

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