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Proof step 32 from proof intshiftrsign-rec-pos (previous step)

abs(i0 + 1) < 32, i0 < 31, 0 ≤ i, 0 ≤ i0, 1 < i, 0 < i ⊦ bits2int ( sublist(0, # int2bits(i / 2), mklist(false, i→n(i0 + 1)) + int2bits(i / 2) + (i % 2 = 1)') + (mklist(false, i→n(i0 + 1)) + int2bits(i / 2) + (i % 2 = 1)')[# int2bits(i / 2)] ') = bits2int(sublist(0, # int2bits(i / 2), mklist(false, i→n(i0 % 32)) + int2bits(i / 2)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0);% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ 0 ≤ i → abs(i) = i;% ⊦ 0 ≤ i → abs(i + 1) = abs(i) + 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ mklist(boolvar, 0) = [];%n ≠ 0 ⊦ mklist(false, (* n)) = false ' + mklist(false, n -1);% ⊦ mklist(boolvar, m + n) = mklist(boolvar, m) + mklist(boolvar, n);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;