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Proof step 24 from proof int2bytes-len-le-neg-02 (previous step)

int2bytes(i) ≠ [], 1 ≤ n, 2 ^ (8 * n→i(n)) + i < 0, 1 ≤ # int2bytes(i), i < 0, 2 ^ (8 * (n→i(# int2bytes(i)) - 1)) + i < 0, # int2bytes(i) ≤ n, 0 ≤ i + 2 ^ ((n→i((# int2bytes(i))-1) + 1) * 8 - 1), 8 * n→i(n) + 1 < (n→i((# int2bytes(i))-1) + 1) * 8, 8 * (n→i(# int2bytes(i)) - 1) + 1 < (n→i((# int2bytes(i))-1) + 1) * 8 ⊦ int2bytes(i).first = i2b(~1), b2i(int2bytes(i).first) = ~1, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ 0 ≤ n→i(n);% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0;% ⊦ i < j - i0 ↔ i + i0 < j;%0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 1 ≤ i ↔ 0 < i;%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ 0 < n ↔ n ≠ 0;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i * j = j * i;% ⊦ i + (j - i) = j;% ⊦ n ≤ m → n→i(m - n) = n→i(m) - n→i(n);% ⊦ n -1 = n - 1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;