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Proof step 23 from proof int2bits-2pot-04 (previous step)

¬(mklist(false, i→n(i)) ≠ [] ∧ (¬ mklist(false, i→n(i)).first) ∈ bitsinvert(uint2bits(~ j)) +1), 0 < i, 0 < 2 ^ i, j < 2 ^ i, j < 0, j + 1 < 2 ^ i, j < i, j * 2 ^ i < i, j + 1 < i ⊦ sgn(bitsinvert(uint2bits(~ j)) +1 + mklist(false, i→n(i))) = int2bits(j) + mklist(false, i→n(i)), j = 0, i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
0 < i, 0 < 2 ^ i, j < 2 ^ i, j < 0, j + 1 < 2 ^ i, j < i, j * 2 ^ i < i, j + 1 < i ⊦ sgn(bitsinvert(uint2bits(~ j)) +1 + mklist(false, i→n(i))) = int2bits(j) + mklist(false, i→n(i)), j = 0, i = 0, (¬ mklist(false, i→n(i)).first) ∈ bitsinvert(uint2bits(~ j)) +1

Rule argument

Additional information

Used simplifier rules: ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ mklist(boolvar, m) = [] ↔ m = 0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;