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Proof step 1 from proof int2bytes-len-eq-pos (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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