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Proof step 28 from proof int2bits-add-low (previous step)

0 < 2 ^ (i - 1) + j0, j0 < 0, j < 0, 0 < i, i < n→i(# int2bits(~(j * 2 ^ i + j0))), j * 2 ^ i + j0 < 0, 0 ≤ (j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i)) * 2 ^ i + j0, j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i) ≤ 0 ⊦ j0 = 0, sgn(uint2bits((j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i)) * 2 ^ i + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0)), j * 2 ^ i + j0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
0 < 2 ^ (i - 1) + j0, j0 < 0, j < 0, 0 < i, i < n→i(# int2bits(~(j * 2 ^ i + j0))), j * 2 ^ i + j0 < 0, 0 ≤ (j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i)) * 2 ^ i + j0, j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i) ≤ 0 ⊦ j0 = 0, sgn(uint2bits((j + 2 ^ (n→i(# int2bits(~(j * 2 ^ i + j0))) - i)) * 2 ^ i + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0)), j * 2 ^ i + j0 = 0

Proof step 28 from proof int2bits-add-low (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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