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Proof step 12 from proof int2bits-add-high (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ i, j * 2 ^ i + j0 < 0 ⊦ sgn(int2bits(2 ^ i * (2 ^ n→i(# int2bits(~ j)) + j) + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0).rest)

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i * j0 = i * (j + j0);