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Proof step 3 from proof abs-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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;%mone-01: ⊦ i + ~1 = i - 1;%irref: ⊦ ¬ i < i; from specification int-basic1%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2