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Proof step 18 from proof int2bytes-butlast (previous step)

i % 256 = 0, i < ~128, ~256 < i, i < 0 ⊦ (i - 255) / 256 = ~1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ i % 256 = 0 ∧ i < 0 → (i - 255) / 256 = i / 256
Proof step 20:
i % 256 = 0 ∧ i < 0, i % 256 = 0, i < ~128, ~256 < i, i < 0 ⊦ i / 256 = ~1

Rule argument

seq: ⊦ i % 256 = 0 ∧ i < 0 → (i - 255) / 256 = i / 256 subst: [i] → [i] rotate: F spec: int-div inst: name: div-minus255-256-01

Additional information

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