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Proof step 18 from proof uint2bits-add-low (previous step)

2 ^ i % 2 = 0 ∧ j ≤ 0 ∧ abs(j) ≤ 2 ^ i, j ≤ 2 ^ (i - 1), j ≤ 2 ^ i, 0 ≤ 2 ^ (i - 1), 1 < j + 2 ^ i, j < 2 ^ (i - 1), j + 1 < i, 0 < 2 ^ (i - 1), j < 0, 0 < i, 0 ≤ j + 2 ^ (i - 1), j < i, j + 1 < 2 ^ (i - 1), j ≤ 0, 0 < j + 2 ^ i, 0 ≤ j + 2 ^ i, j < 2 ^ i, j + 1 < 2 ^ i, 1 < 2 ^ i, 0 < 2 ^ i, 0 ≤ 2 ^ i, ∀ i0, j. i→n(i0) < i→n(i) → true → j < 0 ∧ 0 < i0 ∧ abs(j) ≤ 2 ^ (i0 - 1) → uint2bits(j + 2 ^ i0) = false ' + fillsgn(i→n(i0), int2bits(j)) ⊦ uint2bits(2 ^ (i - 1) + (j - 1) / 2) + (~ j % 2 = 1)' = false ' + fillsgn(i→n(i), int2bits(j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
j ≤ 2 ^ (i - 1), j ≤ 2 ^ i, 0 ≤ 2 ^ (i - 1), 1 < j + 2 ^ i, j < 2 ^ (i - 1), j + 1 < i, 0 < 2 ^ (i - 1), j < 0, 0 < i, 0 ≤ j + 2 ^ (i - 1), j < i, j + 1 < 2 ^ (i - 1), j ≤ 0, 0 < j + 2 ^ i, j < 2 ^ i, j + 1 < 2 ^ i, 1 < 2 ^ i, 0 < 2 ^ i, 0 ≤ 2 ^ i, 0 ≤ j + 2 ^ i, ∀ i0, j. i→n(i0) < i→n(i) → true → j < 0 ∧ 0 < i0 ∧ abs(j) ≤ 2 ^ (i0 - 1) → uint2bits(j + 2 ^ i0) = false ' + fillsgn(i→n(i0), int2bits(j)) ⊦ uint2bits(2 ^ (i - 1) + (j - 1) / 2) + (~1 = j % 2)' = false ' + fillsgn(i→n(i), int2bits(j))

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ ~ i = j ↔ i + j = 0;%j ≠ 0 ⊦ ~ i % j = ~(i % j);% ⊦ 0 < i → 2 ^ i % 2 = 0;