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Proof step 9 from proof int2bits-len-pos-pred (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
0 ≤ i, 0 < j, 2 ^ i < j, j < 2 ^ i * 2 ⊦ 0 = j, # int2bits(j - 1) = i→n(i + 2), 2 ^ i = j, ∃ i. 2 ^ i = j

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;% ⊦ i ≠ j → (i ≤ j ↔ i < j);