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Proof step 34 from proof shiftzero-rec-neg-shorter (previous step)

m < #(mklist(false, m0) + (mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2)) + (i % 2 ≠ 0)'), i→n(j) = m0 + 1, # int2bits((i - 1) / 2) ≤ m, i < ~1, i < 0, 1 < j, 0 < j ⊦ i = ~1, j = 0, bits2uint(firstn(m, (mklist(false, m0) + (mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2)) + (i % 2 ≠ 0)').butlast)) = bits2int(firstn(m, mklist(false, m0) + mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
i→n(j) = m0 + 1, # int2bits((i - 1) / 2) ≤ m, i < ~1, i < 0, 1 < j, 0 < j ⊦ i = ~1, j = 0, bits2uint(firstn(m, mklist(false, m0) + mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2))) = bits2int(firstn(m, mklist(false, m0) + mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2)))

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m + n < m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n ≤ n0 → (m + (n0 - n)) + n = m + n0;% ⊦ # boolvar ' = 0 +1;% ⊦ # mklist(boolvar, n) = n;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;