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Proof step 4 from proof bytes2int-zero-05 (
previous step
)
bytes ≠ [] → (bytes2int(bytes) = 0 ↔ (∀ by. by ∈ bytes → b2i(by) = 0)), bytes2int(bytes) = 0, n < # bytes ⊦ bytes[n] = i2b(0), bytes = []
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 5
:
bytes2int(bytes) = 0, n < # bytes, ∀ by. by ∈ bytes → b2i(by) = 0 ⊦ bytes[n] = i2b(0), bytes = []
Rule argument
Additional information