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Proof step 9 from proof lst-02 (previous step)

i < j + i1 -1, ∀ i, j, j0. i < j + i1 → j0 < j ∨ i < j0 + i1 ⊦ i < j0 + i1 -1, j0 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
i + 1 < j + i1, ∀ i, j, j0. i < j + i1 → j0 < j ∨ i < j0 + i1 ⊦ j0 < j, i + 1 < j0 + i1

Rule argument

Additional information

Used simplifier rules: add-pred: ⊦ i + j -1 = (i + j)-1;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%succ-01: ⊦ i +1 = i + 1;% ⊦ i < j -1 ↔ i +1 < j; (s)