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Proof step 19 from proof shiftzero-rec-neg-longer (previous step)

i→n(j) = m + 1 ∧ i→n(j) ≠ 0, i < ~1, 0 < j, 1 < j, i < 0, n ≤ # int2bits((i - 1) / 2), n ≤ # int2bits((i - 1) / 2) + 1 ⊦ i = ~1, bits2int(firstn(# int2bits((i - 1) / 2) + 1, mklist(false, i→n(j)) + int2bits((i - 1) / 2) + (i % 2 ≠ 0)')) = bits2int(firstn(# int2bits((i - 1) / 2), mklist(false, m) + int2bits((i - 1) / 2)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
i→n(j) = m + 1, i < ~1, 0 < j, 1 < j, i < 0, n ≤ # int2bits((i - 1) / 2), n ≤ # int2bits((i - 1) / 2) + 1 ⊦ i = ~1, bits2int(firstn(# int2bits((i - 1) / 2) + 1, mklist(false, m + 1) + int2bits((i - 1) / 2) + (i % 2 ≠ 0)')) = bits2int(firstn(# int2bits((i - 1) / 2), mklist(false, m) + int2bits((i - 1) / 2))), j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i→n(i) = 0 ↔ i = 0;