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Proof step 26 from proof int2bytes-len-le-neg-01 (previous step)

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

Rule argument

Additional information

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