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Proof step 4 from proof uint2bits-add-high-02 (previous step)

¬(j0 < 0 ∧ 0 < i ∧ 2 ^ (i - 1) < abs(j0) ∧ abs(j0) ≤ 2 ^ i ∧ 1 < j), 0 < j, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 ≤ j0 + 2 ^ i ⊦ uint2bits(j * 2 ^ i + j0) = sgn(uint2bits(j - 1) + fillsgn(i→n(i), int2bits(j0).rest))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ j0 + 2 ^ i, 2 ^ (i - 1) + j0 < 0, 0 < i, j0 < 0 ⊦ uint2bits(2 ^ i + j0) = sgn(false ' + fillsgn(i→n(i), int2bits(j0).rest))

Rule argument

Additional information

Used simplifier rules: uint2bits-zero: ⊦ uint2bits(0) = false ';% ⊦ i = j → i - j = 0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i = 1 → i * j = j;% ⊦ 0 < i → (i ≤ 1 ↔ i = 1);% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i < ~ j ↔ i + j < 0;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;