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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ # bits = 0 ↔ bits = [];% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;