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Proof step 23 from proof int2bytes-first-zerobyte (previous step)

n→i(# int2bits(i)) % 8 = 1, 8 ≤ # int2bits(i), 0 < i, i ≤ 127, ~128 ≤ 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 24:
⊦ 0 < i → (m ≤ # int2bits(i) ↔ ¬ i < 2 ^ (n→i(m) - 2))
Proof step 25:
0 < i, n→i(# int2bits(i)) % 8 = 1, ¬ i < 2 ^ (n→i(8) - 2), 0 < i, i ≤ 127, ~128 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ i = 0

Rule argument

seq: ⊦ 0 < i → (m ≤ # int2bits(i) ↔ ¬ i < 2 ^ (n→i(m) - 2)) subst: [i, m] → [i, 8] rotate: F spec: int2bits inst: name: int2bits-lelen-06

Additional information

Used terms: i, 8