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Proof step 26 from proof int2bytes-len-le-neg-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
int2bytes(i).first = i2b(~1), # int2bytes(i) = 1, 2 ^ (8 * n→i(n) - 1) + i < 0, 1 ≤ n, 0 ≤ b2i(int2bytes(i).rest.first), ~256 ≤ i, 128 + i < b2i(int2bytes(i).rest.first), i < ~128, i < 0 ⊦ n = 0, int2bytes(i).rest = []

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ i + 0 = i;% ⊦ m < 1 ↔ m = 0;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;