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Proof step 55 from proof int2bits-add-high (previous step)

2 ^ n→i(# int2bits(1 - j)) + j = 1, 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 ⊦ sgn(sgn(false ' + fillsgn(i→n(i), int2bits(j0).rest)).rest) = fillsgn(i→n(i), int2bits(j0).rest), ∃ i0. 1 = 2 ^ i0 + j

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: n→i(# int2bits(1 - j))

Additional information

Used terms: n→i(# int2bits(1 - j))