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Proof step 14 from proof int2bits-rec-neg (previous step)

j = i * 2 + j0 ∧ j0 ≤ 0 ∧ abs(j0) < 2 ∧ 0 < 2, 0 = j0, j < 0, j ≤ 0, (j - 1) / 2 < 0 ⊦ ~1 = j, sgn(bitsinvert(uint2bits(~ i)) +1 + false ') = sgn(bitsinvert(uint2bits((1 - j) / 2)) +1) + false '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
(i * 2 - 1) / 2 < 0, i < 0, i ≤ 0 ⊦ sgn(bitsinvert(uint2bits(~ i)) +1 + false ') = sgn(bitsinvert(uint2bits(~ i)) +1) + false '

Rule argument

Additional information

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