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Proof step 15 from proof int2bytes-first-zerobyte-01 (previous step)

# int2bits(j) = i→n(i0 * 8) + i→n(j0), j0 < 8, 0 ≤ j, 0 ≤ j0, 0 ≤ i0 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ 0 ≤ i ∧ 0 ≤ j → i→n(i * j) = i→n(i) * i→n(j)
Proof step 17:
0 ≤ i0 ∧ 0 ≤ 8, # int2bits(j) = i→n(i0) * i→n(8) + i→n(j0), j0 < 8, 0 ≤ j, 0 ≤ j0, 0 ≤ i0 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

Rule argument

seq: ⊦ 0 ≤ i ∧ 0 ≤ j → i→n(i * j) = i→n(i) * i→n(j) subst: [j, i] → [8, i0] rotate: F spec: intnat inst: name: i2n-mult

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false); Used terms: 8, i0