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Proof step 25 from proof intshiftrsign-rec-neg (previous step)

i→n(j) ≠ 0, 0 ≤ j, i < 0, j < 32, 0 < j ⊦ i = ~1, bits2int(sublist(0, # int2bits((i - 1) / 2) + 1, (true + mklist(true, i→n(j) - 1)) + int2bits((i - 1) / 2) + (i % 2 ≠ 0)')) = bits2int(sublist(0, # int2bits((i - 1) / 2), mklist(true, i→n(j - 1)) + int2bits((i - 1) / 2)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
i < 0, j < 32, 0 < j ⊦ i = ~1, bits2int(sublist(0, # int2bits((i - 1) / 2) + 1, (true ' + mklist(true, i→n(j) - 1)) + int2bits((i - 1) / 2) + (i % 2 ≠ 0)')) = bits2int(sublist(0, # int2bits((i - 1) / 2), mklist(true, i→n(j - 1)) + int2bits((i - 1) / 2))), j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ boolvar + bits = boolvar ' + bits;