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Proof step 12 from proof bytes2bytes-length (previous step)

¬ i2b(b2i(by)) ≠ i2b(0), b2i(by) = 0, 1 + # bytes1 < # int2bytes(bytes2int(by ' + bytes1)), ∀ bytes0. # bytes0 < #(by + bytes1) → # bytes0 < # int2bytes(bytes2int(bytes0)) → bytes0 = [] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ b2i(i2b(0)) = 0;% ⊦ i2b(b2i(by)) = by;