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Proof step 30 from proof bytes2bytes-first-neg (previous step)

bytes = by0 + bytes1, b2i(by0) < 0, 0 ≤ b2i(int2bytes(bytes2int(i2b(~1) ' + bytes)).first), ∀ by, bytes0. # bytes0 < # bytes → true → (b2i(int2bytes(bytes2int(by ' + bytes0)).first) < 0 ↔ ¬ 0 ≤ b2i(by)) ⊦ bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
0 ≤ b2i(int2bytes(bytes2int(i2b(~1) ' + by0 ' + bytes1)).first), b2i(by0) < 0, ∀ 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)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' + bytes ≠ [];% ⊦ by + bytes = by ' + bytes;