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Proof step 35 from proof int2bytes-butlast (previous step)

i < 0, i ≤ ~256, i < ~256 ⊦ int2bytes(i / 256 - 1) = int2bytes(i >> 8), i % 256 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 36:
⊦ i < 0 ∧ 8 ≤ j → i >> j = (i - 255) / 256 >> (j - 8)
Proof step 37:
i < 0 ∧ 8 ≤ 8, i < 0, i ≤ ~256, i < ~256 ⊦ int2bytes(i / 256 - 1) = int2bytes((i - 255) / 256 >> (8 - 8)), i % 256 = 0

Rule argument

seq: ⊦ i < 0 ∧ 8 ≤ j → i >> j = (i - 255) / 256 >> (j - 8) subst: [j, i] → [8, i] rotate: F spec: intops inst: name: shiftright-rec-8-neg

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false); Used terms: 8, i