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Proof step 4 from proof int2bytes-len-le-neg-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i < 0, bytes2int(int2bytes(i).first ' + int2bytes(i).rest) + 2 ^ (8 * n→i(# int2bytes(i).rest)) < 0 ⊦ int2bytes(i).first = i2b(~1), # int2bytes(i) ≤ n ↔ ¬ 2 ^ (8 * n→i(n)) + i < 0, b2i(int2bytes(i).first) = ~1

Rule argument

Additional information

Used simplifier rules: bytes2int-first-lez: ⊦ b2i(int2bytes(i).first) < 0 ↔ i < 0;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ b2i(by) = i → (i2b(i) ≠ by ↔ false);% ⊦ i < ~ j ↔ i + j < 0;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;