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Proof step 1 from proof int2int-pos (conclusion)

⊦ 0 ≤ i → bytes2int(int2bytes(i)) = i

User applied rule induction

And obtained the following new goals

Proof step 2:
true, ∀ i0. i→n(i0) < i→n(i) → true → 0 ≤ i0 → bytes2int(int2bytes(i0)) = i0 ⊦ 0 ≤ i → bytes2int(int2bytes(i)) = i

Rule argument

indvar: i→n(i) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(i)