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Proof step 2 from proof int2bytes-first-zerobyte (previous step)

0 ≤ i ⊦ int2bytes(i).first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1

User applied rule induction

And obtained the following new goals

Proof step 3:
0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ int2bytes(i).first = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1

Rule argument

indvar: i→n(i) wfpred: <

Additional information

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