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Proof step 1 from proof int2bits-len-neg (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i < i + j ↔ 0 < j;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i + j < j ↔ i < 0;% ⊦ i +1 = i + 1;% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 1 * i = i;% ⊦ j < (* i1) * j ↔ 0 < (i1 - 1) * j;% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i < ~ j ↔ i + j < 0;% ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;