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Proof step 1 from proof bytes2bytes-first-pos (conclusion)

⊦ 0 ≤ b2i(int2bytes(bytes2int(by ' + bytes)).first) ↔ 0 ≤ b2i(by)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ b2i(int2bytes(bytes2int(by ' + bytes)).first), b2i(by) < 0 ⊦

Rule argument

Additional information

Used simplifier rules: bytes2bytes-first-neg: ⊦ b2i(int2bytes(bytes2int(by ' + bytes)).first) < 0 ↔ b2i(by) < 0;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;