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Proof step 6 from proof int2bits-neg-02 (previous step)

j < 0, j < 0 ⊦ ¬(∃ i. 2 ^ i = ~ j) ∧ 0 ≤ ~ j, sgn(int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest) = int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest, ∃ i. 2 ^ i + j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
j + 2 ^ i = 0, j < 2 ^ i, 0 < 2 ^ i, j < 0, j + 1 < 2 ^ i ⊦ sgn(int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest) = int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest, ∃ i. 2 ^ i + j = 0

Rule argument

Additional information

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