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Proof step 2 from proof int2bytes-first-zerobyte-01 (previous step)

0 ≤ j ⊦ int2bytes(j).first = i2b(0) ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i → (int2bytes(i).first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1)
Proof step 4:
0 ≤ j, 0 ≤ j ⊦ n→i(# int2bits(j)) % 8 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

Rule argument

name: int2bytes-first-zerobyte rotate: F subst: [i] → [j]

Additional information

Used terms: j