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Proof step 26 from proof int2bytes-first-zerobyte (previous step)

n→i(# int2bits(i)) % 8 = 1, 0 < i, i ≤ 127, 64 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
0 ≤ i, 2 ^ i ≤ j, j < 2 ^ (i + 1) ⊦ # int2bits(j) = i→n(i + 2)

Rule argument

seq: 0 ≤ i, 2 ^ i ≤ j, j < 2 ^ (i + 1) ⊦ # int2bits(j) = i→n(i + 2) subst: [i, j] → [6, i] rotate: F spec: int2bits inst: name: int2bits-len-pos

Additional information

Used simplifier rules: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;%j0 ≠ 0, j < j0 ⊦ i ≤ (* j) → i < (* j0); Used terms: 6, i