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Proof step 13 from proof int2bits-restsgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
0 < 2 ^ (i + 2), 0 ≤ 2 ^ (i + 2), j ≤ 2 ^ (i + 2), 0 < 2 ^ i, 0 < j, j < 2 ^ i * 2, 2 ^ i ≤ j, 0 ≤ i ⊦ sgn(uint2bits(2 ^ (i + 2) - j).rest) = uint2bits(2 ^ (i + 2) - j).rest, 0 = j, ∃ i. 2 ^ i = j

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ i → (2 ^ i ≤ 2 ^ j ↔ i ≤ j);% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ i * j = j * i;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ i + j ↔ 0 ≤ j;% ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ i → abs(i) = i;% ⊦ 0 ≤ i ∧ 0 ≤ j → abs(i + j) = abs(i) + abs(j);% ⊦ n→i(i→n(i)) = abs(i);