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Proof step 2 from proof zero (previous step)

i + j = 0, i < 0, j < 0 ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ls-02: ⊦ i < i + j ↔ 0 < j;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;% ⊦ i < j → ¬ j < i; (s)%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1