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Proof step 23 from proof int2bytes-len-ls-01 (previous step)

# int2bits(i) = 1 + i→n(i0 * 8), 0 < i, 0 ≤ i0 ⊦ # int2bits(i) % 8 = 0, i = 0, n ≤ # int2bits(i) / 8 ↔ ¬ i < 2 ^ (n→i(n) * 8 - 1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 24:
⊦ 0 ≤ i ∧ 0 ≤ j → i→n(i * j) = i→n(i) * i→n(j)
Proof step 25:
0 ≤ i0 ∧ 0 ≤ 8, # int2bits(i) = 1 + i→n(i0) * i→n(8), 0 < i, 0 ≤ i0 ⊦ # int2bits(i) % 8 = 0, i = 0, n ≤ # int2bits(i) / 8 ↔ ¬ i < 2 ^ (n→i(n) * 8 - 1)

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