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

m + 1 ≠ 0, 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, (false + mklist(false, m + 1 - 1)) + int2bits((i - 1) / 2) + (i % 2 ≠ 0)')) = bits2int(firstn(# int2bits((i - 1) / 2), mklist(false, m) + int2bits((i - 1) / 2))), j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
i→n(j) = m + 1, i < ~1, 0 < j, 1 < j, i < 0, n ≤ # int2bits((i - 1) / 2), n ≤ # int2bits((i - 1) / 2) + 1 ⊦ j = 0, i = ~1, bits2uint(firstn(# int2bits((i - 1) / 2), mklist(false, m) + int2bits((i - 1) / 2) + (i % 2 ≠ 0)')) = bits2int(firstn(# int2bits((i - 1) / 2), mklist(false, m) + int2bits((i - 1) / 2)))

Rule argument

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0;% ⊦ bits2int(false ' + bits) = bits2uint(bits);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ firstn(n + 1, boolvar ' + bits) = boolvar + firstn(n, bits);% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 + m = m;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;