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Proof step 10 from proof bytes2int-first-pos (previous step)

0 ≤ bytes2int(by ' + bytes.butlast), 0 ≤ b2i(by), band(255, b2i(bytes.last)) + 256 * bytes2int(by ' + bytes.butlast) < 0, ∀ by, bytes0. # bytes0 < # bytes → bytes2int(by ' + bytes0) < 0 ∧ 0 ≤ b2i(by) → false ⊦ bytes2int(by ' + bytes.butlast) = 0, bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ b2i(by), band(255, b2i(bytes.last)) + 256 * bytes2int(by ' + bytes.butlast) < 0, 0 < bytes2int(by ' + bytes.butlast), ∀ by, bytes0. # bytes0 < # bytes → bytes2int(by ' + bytes0) < 0 ∧ 0 ≤ b2i(by) → false ⊦ bytes2int(by ' + bytes.butlast) = 0, bytes = []

Rule argument

Additional information

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