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Proof step 7 from proof uint2bits-len-add (previous step)

fillsgn(i→n(i) + 1, uint2bits(j0)) ≠ [], 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 8:
0 ≤ i, 0 ≤ j0, 0 < j, j0 < 2 ^ i ⊦ fillsgn(i→n(i) + 1, uint2bits(j0)) = [], # fillsgn(i→n(i) + 1, uint2bits(j0)) - 1 = i→n(i)

Rule argument

Additional information

Used simplifier rules: ⊦ n -1 = n - 1;