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Proof step 16 from proof int2bytes-len-pos (previous step)

# int2bits(j) = n * 8, 8 * i < abs(8 * (i + 1) - 1), 0 < j, i→n(8 * i) + 1 < n * 8, 0 ≤ i, n * 8 ≤ i→n(8 * (i + 1) - 1) + 1, 8 * i + 1 < 8 * (i + 1) ⊦ n = 0, n = i→n(i) + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ 0 ≤ i ∧ 0 ≤ j → i→n(i * j) = i→n(i) * i→n(j)
Proof step 18:
0 ≤ 8 ∧ 0 ≤ i, # int2bits(j) = n * 8, 8 * i < abs(8 * (i + 1) - 1), 0 < j, i→n(8) * i→n(i) + 1 < n * 8, 0 ≤ i, n * 8 ≤ i→n(8 * (i + 1) - 1) + 1, 8 * i + 1 < 8 * (i + 1) ⊦ n = 0, n = i→n(i) + 1

Rule argument

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

Additional information

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