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Proof step 70 from proof shiftright-rec-neg (previous step)

0 < abs(j) ∧ abs(j) ≠ 1, 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(true ' + firstn(# int2bits((i - 1) / 2), mklist(true, i→n(j) - 1) + int2bits((i - 1) / 2))) = 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 71:
int2bits(i % 2).butlast = [], i < ~1, i < 0 ⊦ abs(j) = 1, i→n(j) + # int2bits(i % 2) = 0, int2bits((i - 1) / 2) + int2bits(i % 2) = [], j = 0, bits2int(true ' + firstn(# int2bits((i - 1) / 2), mklist(true, i→n(j) - 1) + int2bits((i - 1) / 2))) = bits2int(firstn(# int2bits((i - 1) / 2), mklist(true, i→n(j) - 1) + int2bits((i - 1) / 2)))

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < abs(i) ↔ i ≠ 0;% ⊦ n -1 = n - 1;% ⊦ i→n(abs(i)) = i→n(i);