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Proof step 26 from proof int2bits-add-high (
previous step
)
⊦ j0 < 0 ∧ 0 < i ∧ 2 ^ (i - 1) < abs(j0) ∧ abs(j0) ≤ 2 ^ i ∧ 0 < j → uint2bits(j * 2 ^ i + j0) = sgn(uint2bits(j - 1) + fillsgn(i→n(i), int2bits(j0).rest))
Lemma
uint2bits-add-high-02