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Proof step 4 from proof int2bytes-len-le-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
int2bytes(i).first = i2b(0), ¬ # int2bytes(i) < n ↔ 2 ^ (8 * n→i(n - 1) - 1) ≤ i, 0 ≤ i ⊦ n ≤ # int2bytes(i) ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8 - 1)

Rule argument

Additional information

Used simplifier rules: ⊦ i -1 = i - 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m < n + 1 ↔ ¬ n < m;%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 ≤ m;% ⊦ n -1 = n - 1;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;