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Proof step 64 from proof shiftzero-pos (previous step)

n = n0 + i→n(j) ∧ i→n(j) ≤ n, # int2bits(i) = i→n(j), 0 ≤ i, i→n(j) < n ⊦ bits2int(firstn(n, mklist(false, i→n(j)) + mklist(false, n0) + int2bits(i))) = 0, j = 0, mklist(false, i→n(j)) + mklist(false, n0) = mklist(false, n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 65:
# int2bits(i) = i→n(j), 0 ≤ i ⊦ bits2int(firstn(n0 + i→n(j), mklist(false, i→n(j)) + mklist(false, n0) + int2bits(i))) = 0, j = 0, mklist(false, i→n(j)) + mklist(false, n0) = mklist(false, n0) + mklist(false, i→n(j)), n0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < m + n ↔ n ≠ 0;% ⊦ m < n → m ≤ n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ mklist(boolvar, m + n) = mklist(boolvar, m) + mklist(boolvar, n);