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Proof step 20 from proof int2bytes-len-le-neg-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
int2bytes(i).first = i2b(~1), 2 ^ (8 * (1 + n→i(# int2bytes(i) - 1 - 1)) - 1) + i < 0, 0 ≤ 2 ^ (8 * (1 + n→i(# int2bytes(i) - 1 - 1))) + i, 0 ≤ b2i(int2bytes(i).rest.first) ⊦ n ≤ # int2bytes(i) ↔ ¬ 0 ≤ i + 2 ^ (8 * (n→i(n) - 1) - 1), int2bytes(i).rest = []

Rule argument

Additional information

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