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Proof step 2 from proof minus-one (previous step)

i + 1 = 0 ⊦ i * j = ~ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i + 1 = 0, i < 0 ⊦ j + i * j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ i; (s)% ⊦ i < j + 1 ↔ i ≤ j; (s)%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%mone-03: ⊦ i + 1 = 0 → i < 0; (forward) from specification int-basic2% ⊦ ~ i = j ↔ i + j = 0; (s)