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Proof step 16 from proof uint2bits-len-add (previous step)

0 ≤ j0, i→n(i) + 1 < # int2bits(j0), j0 < 2 ^ i, 0 < j, 0 ≤ j0, 0 ≤ i ⊦ # fillsgn(i→n(i) + 1, int2bits(j0)) - 1 = i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
0 ≤ i, 0 ≤ j0, 0 < j, j0 < 2 ^ i, i→n(i) + 1 < # int2bits(j0) ⊦ # fillsgn(i→n(i) + 1, int2bits(j0)) - 1 = i→n(i)

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)