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Proof step 2 from proof bytes2int-neg (previous step)

true, ∀ bytes0. # bytes0 < # bytes → true → bytes0 ≠ [] → (bytes2int(bytes0) < 0 ↔ b2i(bytes0.first) < 0) ⊦ bytes ≠ [] → (bytes2int(bytes) < 0 ↔ b2i(bytes.first) < 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ bytes0. # bytes0 < # bytes → true → bytes0 ≠ [] → (bytes2int(bytes0) < 0 ↔ b2i(bytes0.first) < 0) ⊦ bytes = [], bytes2int(bytes) < 0 ↔ ¬ 0 ≤ b2i(bytes.first)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;