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Proof step 72 from proof bib-neg-high (previous step)

i < ~256, 8 ≤ # int2bits(i), lastn(8, int2bits(i)).first, i < 0, ~128 ≤ i % 256 ⊦ ~256 = i, b2i(i2b(i % 256)) = i % 256

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: byte inst: name: num-ibbi seq: ~128 ≤ i, i ≤ 127 ⊦ b2i(i2b((* i))) = i subst: [i] → [i % 256] flags: rewrite, current

Additional information

Used simplifier rules: numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%mod-grz: ⊦ 0 < j ∧ i < 0 → ¬ 0 < i % j; from specification int-div Used terms: i % 256