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Proof step 6 from proof int2bytes-len-le-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
int2bytes(i).first = i2b(0), 2 ^ (8 * n→i(n - 1) - 1) ≤ i, 0 ≤ i, n ≤ # int2bytes(i), i < 2 ^ ((n→i(n) - 1) * 8 - 1), 2 ^ (8 * n→i(n - 1) - 1) < 2 ^ ((n→i(n) - 1) * 8 - 1) ⊦ n = 1, n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ i < j - i0 ↔ i + i0 < j;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ m ≤ n → (n < m ↔ false);