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Proof step 7 from proof bytes2int-zero-06 (previous step)

∀ by. by ∈ bytes → b2i(by) = 0, ∀ bytes0. # bytes0 < # bytes → (∀ by. by ∈ bytes0 → b2i(by) = 0) → bytes2int(bytes0) = 0 ∨ bytes0 = [] ⊦ bytes2int(bytes) = 0 ∨ bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ by. by ∈ bytes → b2i(by) = 0, ∀ bytes0. # bytes0 < # bytes → (∀ by. by ∈ bytes0 → b2i(by) = 0) → bytes2int(bytes0) = 0 ∨ bytes0 = [] ⊦ bytes = [], bytes2int(bytes) = 0

Proof step 7 from proof bytes2int-zero-06 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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