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Proof step 15 from proof int2bits-len-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
0 ≤ i, 2 ^ (i + 2) < 2 ^ i * 2 ⊦ 0 = 2 ^ (i + 2)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ i + j ↔ 0 ≤ j;% ⊦ 0 ≤ i → (2 ^ i ≤ 2 ^ j ↔ i ≤ j);%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ i → abs(i) = i;% ⊦ 0 ≤ i ∧ 0 ≤ j → abs(i + j) = abs(i) + abs(j);% ⊦ n→i(i→n(i)) = abs(i);