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Proof step 3 from proof absone-01 (previous step)

i < 0, abs(i) = 2 + j, j < 0 ⊦ j = ~1, i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
~2 = j + i, j < 0, i < 0 ⊦ j = ~1, i = 0

Rule argument

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j;%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2