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Proof step 79 from proof int2bits-add-high (previous step)

int2bits(j0).rest ≠ [], 2 ^ n→i(# int2bits(1 - j)) + j = 2 ^ i0, 0 < i0, j * 2 ^ i + j0 < 0, 0 ≤ i, 0 < 2 ^ i + j0, 2 ^ (i - 1) + j0 < 0, 0 < i, j0 < 0, j < 0, 0 ≤ j0 + 2 ^ i, 0 ≤ 2 ^ i * 2 ^ i0 + j0, int2bits(j0).rest.first ⊦ i0 = 0, sgn(true ' + fillsgn(i→n(i), int2bits(j0).rest)) = true ' + fillsgn(i→n(i), int2bits(j0).rest), ∃ i0. 1 = 2 ^ i0 + j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 80:
2 ^ n→i(# int2bits(1 - j)) + j = 2 ^ i0, int2bits(j0).rest.first, 0 ≤ 2 ^ i * 2 ^ i0 + j0, 0 ≤ j0 + 2 ^ i, j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ i, j * 2 ^ i + j0 < 0, 0 < i0 ⊦ int2bits(j0).rest = [], sgn(true ' + fillsgn(i→n(i), int2bits(j0).rest)) = true ' + fillsgn(i→n(i), int2bits(j0).rest), i0 = 0, ∃ i0. 1 = 2 ^ i0 + j

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;