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Proof step 6 from proof uint2bits-add-high (previous step)

2 ^ (i - 1) + j < 2 ^ i, j + 1 < i, 0 < 2 ^ i, 2 ^ (i - 1) + j < i, (2 ^ (i - 1) + j) + 1 < i, 2 ^ (i - 1) < 2 ^ i, j < 0, 0 < i, 2 ^ (i - 1) + j < 0, 0 ≤ j + 2 ^ i, j < i, j + 1 < 2 ^ i, j < 2 ^ i, (2 ^ (i - 1) + j) + 1 < 2 ^ i, ∀ i0, j. i→n(i0) < i→n(i) → true → j < 0 ∧ 0 < i0 ∧ 2 ^ (i0 - 1) < abs(j) ∧ abs(j) ≤ 2 ^ i0 → uint2bits(j + 2 ^ i0) = sgn(int2bits(j).rest) ⊦ uint2bits(j + 2 ^ i) = sgn(int2bits(j).rest), j + 2 ^ i ≠ 1 ∧ 0 ≠ j + 2 ^ i

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

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