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Proof step 25 from proof int2bytes-len-le-neg-02 (previous step)

0 ≤ 2 ^ (8 * n→i(# int2bytes(i)) - 1) + i, 8 * (n→i(# int2bytes(i)) - 1) + 1 < 8 * n→i(# int2bytes(i)), 8 * n→i(n) + 1 < 8 * n→i(# int2bytes(i)), # int2bytes(i) ≤ n, 2 ^ (8 * (n→i(# int2bytes(i)) - 1)) + i < 0, i < 0, 1 ≤ # int2bytes(i), 2 ^ (8 * n→i(n)) + i < 0, 1 ≤ n ⊦ n = 0, b2i(int2bytes(i).first) = ~1, int2bytes(i).first = i2b(~1)

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

seq: j * i + j0 < j * i0, abs(j0) < j ⊦ i ≤ i0 subst: [j, i, j0, i0] → [8, n→i(n) - 1, 1, n→i(# int2bytes(i))] rotate: F spec: int-mult inst: name: modlsp-02

Additional information

Used simplifier rules: ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ 0 < i → (i * j < i * j0 ↔ j < j0);% ⊦ i < j - i0 ↔ i + i0 < j;% ⊦ i ≥ 0 → ((* i) + n→i(n) < n→i(m) ↔ i→n(i) + n < m);% ⊦ ¬ m + n < m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ i * j * j0 = (i * j) * j0; Used terms: 8, n→i(n) - 1, 1, n→i(# int2bytes(i))