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Proof step 29 from proof int2bytes-first-zerobyte (previous step)

# int2bits(i) = 1, ~128 ≤ i, i ≤ 127, 0 < 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 30:
⊦ 0 ≤ i → (# int2bits(i) = 1 ↔ i = 0)

Rule argument

seq: ⊦ 0 ≤ i → (# int2bits(i) = 1 ↔ i = 0) subst: [i] → [i] rotate: F spec: int2bits inst: name: int2bits-len-one

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i; Used terms: i