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Proof step 7 from proof bytes2bytes-first-neg (previous step)

¬(by ' + bytes ≠ [] ∧ b2i((by ' + bytes).first) ≠ 0 ∧ b2i((by ' + bytes).first) ≠ ~1), ∀ by, bytes0. # bytes0 < # bytes → true → (b2i(int2bytes(bytes2int(by ' + bytes0)).first) < 0 ↔ ¬ 0 ≤ b2i(by)) ⊦ bytes = [], b2i(int2bytes(bytes2int(by ' + bytes)).first) < 0 ↔ ¬ 0 ≤ b2i(by)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ by, bytes0. # bytes0 < # bytes → true → (b2i(int2bytes(bytes2int(by ' + bytes0)).first) < 0 ↔ ¬ 0 ≤ b2i(by)) ⊦ bytes = [], b2i(int2bytes(bytes2int(by ' + bytes)).first) < 0 ↔ ¬ 0 ≤ b2i(by), b2i(by) ≠ 0 ∧ b2i(by) ≠ ~1

Rule argument

Additional information

Used simplifier rules: ⊦ (by ' + bytes).first = by;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' ≠ [];% ⊦ ¬ bytes + bytes0 ≠ [] ↔ bytes = [] ∧ bytes0 = [];