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Proof step 18 from proof int2bytes-len-le-neg-02 (previous step)

b2i(int2bytes(i).first) < 0 → ~(2 ^ ((n→i(# int2bytes(i).rest) + 1) * 8 - 1)) ≤ bytes2int(int2bytes(i).first ' + int2bytes(i).rest), 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 ⊦ int2bytes(i).first = i2b(~1), b2i(int2bytes(i).first) = ~1, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
0 ≤ 2 ^ ((n→i(# int2bytes(i).rest) + 1) * 8 - 1) + bytes2int(int2bytes(i).first ' + int2bytes(i).rest), # int2bytes(i) ≤ n, 2 ^ (8 * (n→i(# int2bytes(i)) - 1)) + i < 0, i < 0, 1 ≤ # int2bytes(i), 2 ^ (8 * n→i(n)) + i < 0, 1 ≤ n ⊦ n = 0, b2i(int2bytes(i).first) = ~1, int2bytes(i).first = i2b(~1)

Rule argument

Additional information

Used simplifier rules: bytes2int-first-lez: ⊦ b2i(int2bytes(i).first) < 0 ↔ i < 0;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;