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Proof step 4 from proof uint2bits-len-add (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i, 0 ≤ j0, 0 < j, j0 < 2 ^ i ⊦ # fillsgn(i→n(i) + 1, uint2bits(j0)).rest = i→n(i)

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;