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Proof step 1 from proof int2bytes-len-ls-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
int2bytes(i).first = i2b(0), 0 ≤ i ⊦ n < # int2bytes(i) ↔ ¬ i < 2 ^ (n→i(n) * 8 - 1)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i -1 = i - 1;