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Proof step 22 from proof int2bytes-len-pos (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ 0 ≤ i ∧ 0 ≤ j → i→n(i * j) = i→n(i) * i→n(j) subst: [j, i] → [i + 1, 8] rotate: F spec: intnat inst: name: i2n-mult

Additional information

Used simplifier rules: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;% ⊦ m * (n + 1) = m * n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ m * n = n * m;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ m * n ≤ m1 * n + n ↔ ¬ ¬(m ≤ m1 + 1 ∨ n = 0);% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); Used terms: i + 1, 8