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Proof step 2 from proof uint2bits-add-low (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
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), ∀ 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(j + 2 ^ i) = false ' + fillsgn(i→n(i), int2bits(j))

Rule argument

Additional information

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