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Proof step 22 from proof uint2bits-add-low (previous step)

j ≠ 0 ∧ i ≠ 1 ∧ ¬ 2 ^ (i - 2) < abs(j - 1) / 2, 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, j ≠ 0 ∧ i ≠ 1 ∧ ¬ 2 ^ (i - 2) < abs(j - 1) / 2 → uint2bits((j - 1) / 2 + 2 ^ (i - 1)) = false ' + fillsgn(i→n(i) - 1, int2bits((j - 1) / 2)), ∀ 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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
uint2bits((j - 1) / 2 + 2 ^ (i - 1)) = false ' + fillsgn(i→n(i) - 1, int2bits((j - 1) / 2)), 0 < j + 2 ^ i, j < 2 ^ i, j + 1 < 2 ^ i, 1 < 2 ^ i, 0 < 2 ^ i, 0 ≤ 2 ^ i, 0 ≤ 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), 0 < i, 0 ≤ j + 2 ^ (i - 1), j < i, j + 1 < 2 ^ (i - 1), j < 0, abs(j - 1) / 2 ≤ 2 ^ (i - 2), false ∈ uint2bits((j - 1) / 2 + 2 ^ (i - 1)), ∀ 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)), i = 1, j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;