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Proof step 22 from proof shiftzero-rec-neg-shorter (previous step)

i→n(j) = m0 + 1 ∧ i→n(j) ≠ 0, # int2bits((i - 1) / 2) ≤ m, i < ~1, # int2bits(i) ≤ m + 1, i < 0, 1 < j, 0 < j ⊦ bits2int(firstn(m + 1, mklist(false, i→n(j)) + fillsgn(m + 1, int2bits(i)))) = bits2int(firstn(m, mklist(false, m0) + fillsgn(m, int2bits((i - 1) / 2))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
i→n(j) = m0 + 1, # int2bits((i - 1) / 2) ≤ m, i < ~1, # int2bits(i) ≤ m + 1, i < 0, 1 < j, 0 < j ⊦ bits2int(firstn(m + 1, mklist(false, m0 + 1) + fillsgn(m + 1, int2bits(i)))) = bits2int(firstn(m, mklist(false, m0) + fillsgn(m, int2bits((i - 1) / 2)))), j = 0

Rule argument

Additional information

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