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Proof step 57 from proof shiftright-rec-neg (previous step)

int2bits(i % 2).butlast = [], i < ~1, i < 0 ⊦ i→n(j) + # int2bits(i % 2) = 0, int2bits((i - 1) / 2) + int2bits(i % 2) = [], j = 0, bits2int ((mklist(true, i→n(j)) + int2bits((i - 1) / 2)).first + firstn(# int2bits((i - 1) / 2), (mklist(true, i→n(j)) + int2bits((i - 1) / 2)).rest)) = bits2int(firstn(# int2bits((i - 1) / 2), mklist(true, i→n(abs(j) - 1)) + int2bits((i - 1) / 2)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 58:
int2bits(i % 2).butlast = [], i < ~1, i < 0 ⊦ i→n(j) + # int2bits(i % 2) = 0, int2bits((i - 1) / 2) + int2bits(i % 2) = [], j = 0, bits2int ((mklist(true, i→n(j)) + int2bits((i - 1) / 2)).first ' + firstn(# int2bits((i - 1) / 2), (mklist(true, i→n(j)) + int2bits((i - 1) / 2)).rest)) = bits2int(firstn(# int2bits((i - 1) / 2), mklist(true, i→n(abs(j) - 1)) + int2bits((i - 1) / 2)))

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar + bits = boolvar ' + bits;