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Proof step 23 from proof int2bytes-len-le-neg-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
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

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ m < n → n ≠ 0; (forward)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;