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Proof step 16 from proof int2bytes-len-neg (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ 0 ≤ i → (2 ^ i ≤ 2 ^ j ↔ i ≤ j)
Proof step 18:
0 ≤ 8 * i, 2 ^ (8 * i) ≤ 2 ^ (8 * (i + 1) - 1), 0 ≤ i, 0 ≤ 2 ^ (8 * (i + 1) - 1) + j, j + 2 ^ (8 * i) ≤ 0, n→i(n0) ≤ i + 1, j + 2 ^ (n→i(n0) * 8 - 2) < 0, 0 ≤ 2 ^ (n→i(n0) * 8 - 1) + j, j < 0, 8 * i ≤ n→i(n0) * 8 - 1 ⊦ n0 = 0, n0 * 8 = 1, n0 = i→n(i) + 1

Rule argument

seq: ⊦ 0 ≤ i → (2 ^ i ≤ 2 ^ j ↔ i ≤ j) subst: [j, i] → [n→i(n0) * 8 - 1, 8 * i] rotate: F spec: int-pot inst: name: two-le

Additional information

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