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Proof step 10 from proof shiftright-rec-neg (previous step)

¬ i < ~1, i < 0 ⊦ j = 0, bits2int(sublist(0, # int2bits(i), mklist(true, i→n(j)) + int2bits(i))) = bits2int(sublist(0, # 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 11:
⊦ j = 0, bits2int(sublist(0, # int2bits(~1), mklist(true, i→n(j)) + int2bits(~1))) = bits2int(sublist(0, # int2bits(~1), mklist(true, i→n(abs(j) - 1)) + int2bits(~1)))

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ ¬ i < j ↔ j ≤ i;