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Proof step 1 from proof int2bytes-first-zerobyte-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;