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Proof step 10 from proof uint2bits-len-add (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ i, 0 ≤ j0, 0 < j, j0 < 2 ^ 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) ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);