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Proof step 7 from proof int2bytes-len-eq-neg-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ ¬ m ≤ n ↔ n < m;