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Proof step 1 from proof uint2bits-len-min (conclusion)

0 ≤ i, 0 ≤ j, 2 ^ i ≤ j ⊦ i→n(i + 1) < # uint2bits(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i, 0 ≤ j, 2 ^ i ≤ j, # uint2bits(j) ≤ i→n(i) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;