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Proof step 20 from proof int2bytes-first-zerobyte (previous step)

n→i(# int2bits(i)) % 8 = 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 21:
⊦ abs(i) < j → i % j = i
Proof step 22:
¬ abs(n→i(# int2bits(i))) < 8, n→i(# int2bits(i)) % 8 = 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
Proof step 28:
abs(n→i(# int2bits(i))) < 8, n→i(# 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

Rule argument

seq: ⊦ abs(i) < j → i % j = i subst: [j, i] → [8, n→i(# int2bits(i))] rotate: F spec: int-div inst: name: mod-zero-05

Additional information

Used terms: 8, n→i(# int2bits(i))