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Proof step 13 from proof uint2bits-len-add (previous step)

¬ # uint2bits(j0) ≤ i→n(i) + 1, 0 ≤ i, 0 ≤ j0, 0 < j, j0 < 2 ^ i ⊦ # fillsgn(i→n(i) + 1, uint2bits(j0)) - 1 = i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i→n(i) + 1 < # uint2bits(j0), j0 < 2 ^ i, 0 < j, 0 ≤ j0, 0 ≤ i ⊦ # fillsgn(i→n(i) + 1, uint2bits(j0)) - 1 = i→n(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 = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;