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Proof step 106 from proof uint2bits-add-high (previous step)

(2 ^ (i - 1) * 2 + 1) / 2 = 2 ^ (i - 1), 1 < 2 ^ i, 0 < 2 ^ i, 0 ≤ 2 ^ (i - 1) * 2 + 2 ^ i, 0 < 2 ^ (i - 1) * 2 + 2 ^ i, 0 ≤ 2 ^ (i - 1) + i, 1 < 2 ^ (i - 1) + 2 ^ (i - 1), 2 ^ (i - 1) + 2 ^ (i - 1) < 2 ^ i, 0 < 2 ^ (i - 1), 2 ^ (i - 1) ≤ 2 ^ i, 0 < i, 2 * (1 - 2 ^ (i - 1)) < i, 2 ^ (i - 1) + 1 < 2 ^ i, 2 ^ (i - 1) * 2 < 2 ^ i, 0 < i + 2 ^ (i - 1) * 2, 1 < 2 ^ (i - 1) * 2 + 2 ^ i, 0 < 2 ^ i + 2 ^ (i - 1) * 2, 0 ≤ 2 ^ i, 2 ^ (i - 1) * 2 ≤ 2 ^ i, 2 ^ (i - 1) + 2 * (1 - 2 ^ (i - 1)) < i, 0 ≤ i, 1 < 2 ^ (i - 1) + i, 0 ≤ 2 ^ (i - 1) + 2 ^ (i - 1), 1 < 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) ⊦ 2 ^ (i - 1) = 0, 2 ^ i = 1 + 2 ^ (i - 1) * 2, i = 1

User applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < i;%0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0;%0 < i ⊦ 2 ^ (i - 1) + 2 ^ (i - 1) = 2 ^ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;