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Proof step 71 from proof int2bytes-len-ls-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 72:
⊦ 0 ≤ i ∧ 0 ≤ j → i→n(i * j) = i→n(i) * i→n(j)

Rule argument

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

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n);% ⊦ i→n(n→i(n)) = n;% ⊦ m * n = n * m;% ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n;% ⊦ m ≤ n → (n < m ↔ false);% ⊦ m0 * m * n = (m0 * m) * n; Used terms: 8, n→i(n)