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Proof step 5 from proof int2bits-lelen (previous step)

0 < i → (∃ i0. 0 ≤ i0 ∧ 2 ^ i0 ≤ i ∧ i < 2 ^ (i0 + 1)), n→i(m) = i0 + 1, 0 < i ⊦ # int2bits(i) ≤ m ↔ ¬ 2 ^ i0 ≤ i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n→i(m) = i0 + 1, i < 2 ^ i1 * 2, 2 ^ i1 ≤ i, 0 ≤ i1, 0 < 2 ^ i1, 0 < i ⊦ # int2bits(i) ≤ m ↔ ¬ 2 ^ i0 ≤ i

Rule argument

Additional information

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