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Proof step 29 from proof shiftzero-rec-neg-longer (previous step)

¬ ( ¬ firstn(# int2bits((i - 1) / 2), mklist(false, m) + int2bits((i - 1) / 2)).first ∧ firstn(# int2bits((i - 1) / 2), mklist(false, m) + int2bits((i - 1) / 2)) ≠ []), i→n(j) = m + 1, i < ~1, 0 < j, 1 < j, i < 0, n ≤ # int2bits((i - 1) / 2), n ≤ # int2bits((i - 1) / 2) + 1 ⊦ m = 0, j = 0, i = ~1, bits2uint(firstn(# int2bits((i - 1) / 2), mklist(false, m) + int2bits((i - 1) / 2))) = 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 30:
i→n(j) = m + 1, i < ~1, 0 < j, 1 < j, i < 0, n ≤ # int2bits((i - 1) / 2), n ≤ # int2bits((i - 1) / 2) + 1, firstn(# int2bits((i - 1) / 2), mklist(false, m) + int2bits((i - 1) / 2)).first ⊦ m = 0, j = 0, i = ~1, bits2uint(firstn(# int2bits((i - 1) / 2), mklist(false, m) + int2bits((i - 1) / 2))) = bits2int(firstn(# int2bits((i - 1) / 2), mklist(false, m) + int2bits((i - 1) / 2)))

Rule argument

Additional information

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