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Proof step 16 from proof bytes2int-neg (previous step)

¬ 0 ≤ b2i(by), ¬ 0 ≤ bytes2int(by ' + bytes) ↔ b2i(by) < 0, ∀ bytes0. # bytes0 < #(by + (bytes + by0 ')) → true → bytes0 ≠ [] → (bytes2int(bytes0) < 0 ↔ b2i(bytes0.first) < 0) ⊦ band(255, b2i(by0)) + 256 * bytes2int(by ' + bytes) < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
bytes2int(by ' + bytes) < 0, 0 ≤ band(255, b2i(by0)) + 256 * bytes2int(by ' + bytes), b2i(by) < 0, ∀ bytes0. # bytes0 < #(by + (bytes + by0 ')) → true → bytes0 ≠ [] → (bytes2int(bytes0) < 0 ↔ b2i(bytes0.first) < 0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ i < j ↔ j ≤ i;