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Proof step 3 from proof plst-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i + i0 < j + j0, j0 < j1 ⊦ i + i1 < j + j1, i0 < i1

Rule argument

Additional information

Used simplifier rules: ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1