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Proof step 29 from proof int2bits-lslen-07 (previous step)

j < 2 ^ i * 2, i→n(i + 2) + 1 < # int2bits(j + 1), 2 ^ i ≤ j, 0 < j, 0 ≤ i, j + 1 < 2 ^ i ⊦ 0 = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
0 ≤ i, 2 ^ i ≤ j, j < 2 ^ (i + 1) ⊦ # int2bits(j) = i→n(i + 2)
Proof step 31:
¬(0 ≤ i -1 ∧ 2 ^ i -1 ≤ j + 1 ∧ j + 1 < 2 ^ (i -1 + 1)), j < 2 ^ i * 2, i→n(i + 2) + 1 < # int2bits(j + 1), 2 ^ i ≤ j, 0 < j, 0 ≤ i, j + 1 < 2 ^ i ⊦ 0 = j

Rule argument

name: int2bits-len-pos rotate: F subst: [i, j] → [i -1, j + 1]

Additional information

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