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Proof step 7 from proof bytes2int-neg (previous step)

1 < #(by ' + bytes1), ∀ bytes0. # bytes0 < #(by + bytes1) → true → bytes0 ≠ [] → (bytes2int(bytes0) < 0 ↔ b2i(bytes0.first) < 0) ⊦ bytes2int((by ' + bytes1).butlast) * 256 + band(b2i((by ' + bytes1).last), 255) < 0 ↔ ¬ 0 ≤ b2i(by)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ bytes0. # bytes0 < #(by + bytes1) → true → bytes0 ≠ [] → (bytes2int(bytes0) < 0 ↔ b2i(bytes0.first) < 0) ⊦ bytes1 = [], band(255, b2i(bytes1.last)) + 256 * bytes2int(by ' + bytes1.butlast) < 0 ↔ ¬ 0 ≤ b2i(by)

Rule argument

Additional information

Used simplifier rules: ⊦ bytes ≠ [] → (bytes0 + bytes).butlast = bytes0 + bytes.butlast;% ⊦ bytes ≠ [] → (by ' + bytes).last = bytes.last;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;