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Proof step 5 from proof uint2bits-add-high-01 (previous step)

⊦ 0 ≤ i ∧ 0 < j ∧ 0 ≤ j0 ∧ j0 < 2 ^ i → uint2bits(j * 2 ^ i + j0) = uint2bits(j) + fillsgn(i→n(i) + 1, uint2bits(j0)).rest

Lemma uint2bits-mod2pot-01