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Proof step 61 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(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)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 62:
⊦ 0 < i ∧ i ≠ 1 → i→n(i - 1) = i→n(i) -1
Proof step 63:
¬(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)))
Proof step 70:
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)))

Rule argument

seq: ⊦ 0 < i ∧ i ≠ 1 → i→n(i - 1) = i→n(i) -1 subst: [i] → [abs(j)] rotate: F spec: intnat inst: name: abs-succ-02

Additional information

Used terms: abs(j)