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Proof step 14 from proof bytes2int-zero-06 (previous step)

1 < # bytes, ∀ by. by ∈ bytes → b2i(by) = 0, ∀ bytes0. # bytes0 < # bytes → (∀ by. by ∈ bytes0 → b2i(by) = 0) → bytes2int(bytes0) = 0 ∨ bytes0 = [] ⊦ bytes = [], bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
∀ by. by ∈ bytes → b2i(by) = 0, ∀ bytes0. # bytes0 < # bytes → (∀ by. by ∈ bytes0 → b2i(by) = 0) → bytes2int(bytes0) = 0 ∨ bytes0 = [] ⊦ bytes = [], bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255) = 0, # bytes = 1

Rule argument

Additional information

Used simplifier rules: ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);