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Proof step 30 from proof int2int-neg (previous step)

0 ≤ i ∧ 2 ^ i ≤ ~ j ∧ ~ j < 2 ^ (i + 1), 2 ^ i + j ≤ n→i(i→n(i + 2)), 2 ^ i + j < n→i(i→n(i + 2)), j < n→i(i→n(i + 2)), j ≤ 0, j < 0, j < 2 ^ i, j + 1 < 2 ^ i, j + 1 < n→i(i→n(i + 2)), 0 < 2 ^ i * 2 + j, 0 ≤ i, 2 ^ i + j ≤ 0, j + 1 < 2 ^ n→i(i→n(i + 2)), 0 ≤ 2 ^ n→i(i→n(i + 2)) + j, 0 < 2 ^ n→i(i→n(i + 2)) + j, 2 ^ i + j ≤ i, 0 < 2 ^ i, j < i, j ≤ i, j ≤ 2 ^ n→i(i→n(i + 2)), 0 ≤ 2 ^ n→i(i→n(i + 2)), j < 2 ^ n→i(i→n(i + 2)), 0 < 2 ^ n→i(i→n(i + 2)), 1 < 2 ^ n→i(i→n(i + 2)), 2 ^ i + j ≤ 2 ^ n→i(i→n(i + 2)), i ≤ n→i(i→n(i + 2)), 2 ^ i + j < 2 ^ n→i(i→n(i + 2)), i < n→i(i→n(i + 2)), j ≤ n→i(i→n(i + 2)) ⊦ 2 ^ n→i(i→n(i + 2)) = 2 ^ (n→i(# uint2bits(2 ^ n→i(i→n(i + 2)) + j)) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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