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Proof step 20 from proof bytes2bytes-first-neg (previous step)

0 ≤ b2i(by0), b2i(int2bytes(bytes2int(by0 ' + bytes1)).first) < 0, 0 ≤ b2i(by0), ∀ by, bytes0. # bytes0 < #(by0 + bytes1) → true → (b2i(int2bytes(bytes2int(by ' + bytes0)).first) < 0 ↔ ¬ 0 ≤ b2i(by)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
b2i(int2bytes(bytes2int(by0 ' + bytes1)).first) < 0, 0 ≤ b2i(by0), ∀ by, bytes0. # bytes0 < #(by0 + bytes1) → true → (b2i(int2bytes(bytes2int(by ' + bytes0)).first) < 0 ↔ ¬ 0 ≤ b2i(by)) ⊦

Rule argument

Additional information

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