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Proof step 21 from proof int2bytes-len-neg (previous step)

n→i(n0) = i + 1, 0 ≤ 2 ^ ((i + 1) * 8 - 1) + j, j < 0, 2 ^ (8 * i) ≤ 2 ^ (8 * (i + 1) - 1), 0 ≤ i, j + 2 ^ (8 * i) ≤ 0, j + 2 ^ ((i + 1) * 8 - 2) < 0 ⊦ n0 = 0, n0 * 8 = 1, n0 = i→n(i) + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: 0 ≤ i ⊦ n→i(n) = (* i) ↔ n = i→n(i) subst: [i, n] → [i + 1, n0] rotate: F spec: intnat inst: name: num-n2i-eq

Additional information

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