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Proof step 2 from proof ls (previous step)

∀ i. i + j0 < j0 ↔ i < 0 ⊦ i + j0 +1 < j0 +1 ↔ i < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ i. i + j0 < j0 ↔ i < 0 ⊦ i + j0 < j0 ↔ i < 0

Rule argument

Additional information

Used simplifier rules: add-succ: ⊦ i + j +1 = (i + j)+1;%add-a: ⊦ i + j + i0 = (i + j) + i0;%ls-succ-02: ⊦ i +1 < j +1 ↔ i < j; from specification int-basic1