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Proof step 6 from proof plst-01 (previous step)

j0 < j1, i + i0 < j + j0 ⊦ i0 = i1, i + i1 < j + j1, i0 < i1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i + i0 < j + j0, j0 < j1, i1 < i0 ⊦ i + i1 < j + j1

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → ¬ j < i; (s)% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)