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Proof step 7 from proof shiftright-rec-neg (previous step)

i < 0 ⊦ j = 0, bits2int(sublist(0, # int2bits(i), mklist(true, i→n(j)) + int2bits(i))) = bits2int(sublist(0, # int2bits((i - 1) / 2), mklist(int2bits((i - 1) / 2).first, i→n(abs(j) - 1)) + int2bits((i - 1) / 2)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
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)))

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ ~1 ↔ i < 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i + (j - i0) = i + j - i0;%0 < j ⊦ i / (* j) < 0 ↔ i + j ≤ 0;% ⊦ int2bits(j).first ↔ j < 0;