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Proof step 3 from proof int2bytes-first-zerobyte (previous step)

0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ int2bytes(i).first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
255 < i ⊦ int2bytes(i) = int2bytes(i / 256) + int2bytes(i % 256).last '
Proof step 5:
¬ 255 < i, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ int2bytes(i).first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1
Proof step 31:
255 < i, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ (int2bytes(i / 256) + int2bytes(i % 256).last ').first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1

Rule argument

name: int2bytes-rec-pos rotate: F subst: [i] → [i]

Additional information

Used terms: i