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Proof step 38 from proof shiftzero-rec-neg-shorter (previous step)

¬ ( ¬ firstn(m, mklist(false, m0) + mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2)).first ∧ firstn(m, mklist(false, m0) + mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2)) ≠ []), i→n(j) = m0 + 1, # int2bits((i - 1) / 2) ≤ m, i < ~1, i < 0, 1 < j, 0 < j ⊦ m0 = 0, i = ~1, j = 0, bits2uint(firstn(m, mklist(false, m0) + mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2))) = bits2int(firstn(m, mklist(false, m0) + mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
i→n(j) = m0 + 1, # int2bits((i - 1) / 2) ≤ m, i < ~1, i < 0, 1 < j, 0 < j, firstn(m, mklist(false, m0) + mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2)).first ⊦ m0 = 0, i = ~1, j = 0, bits2uint(firstn(m, mklist(false, m0) + mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2))) = bits2int(firstn(m, mklist(false, m0) + mklist(true, m - # int2bits((i - 1) / 2)) + int2bits((i - 1) / 2))), m = 0

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m ≤ 0 ↔ m = 0;% ⊦ bits2uint([]) = 0;% ⊦ firstn(0, bits) = [];% ⊦ firstn(n, bits) = [] ↔ n = 0;