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Proof step 64 from proof shiftright-rec-neg (previous step)

int2bits(i % 2).butlast = [], abs(j) = 1, 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))) = (i - 1) / 2

User applied rule cut formula

And obtained the following new goals

Proof step 65:
i→n(j) = i→n(abs(j)), int2bits(i % 2).butlast = [], abs(j) = 1, 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))) = (i - 1) / 2

Rule argument

fma: i→n(j) = i→n(abs(j))

Additional information

Used simplifier rules: ⊦ i→n(abs(i)) = i→n(i);% ⊦ i→n(i) = i→n(j) ↔ abs(i) = abs(j); Used formulas: i→n(j) = i→n(abs(j))