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Proof step 31 from proof int2bytes-len-le-neg-02 (previous step)

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

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, n→i(# int2bytes(i))] rotate: F spec: int-mult inst: name: modlsp-02

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ n→i(m) ≤ n→i(n) ↔ m ≤ n;% ⊦ m ≤ n → (n ≤ m ↔ m = n);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ (* i2) * i + j < (* i1) * i ↔ j < (i1 - i2) * i;% ⊦ i * 0 = 0;% ⊦ i * j * j0 = (i * j) * j0; Used terms: 8, n→i(n), 1, n→i(# int2bytes(i))