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Proof step 4 from proof pls (previous step)

∀ j, i0, j0. i1 < j ∧ i0 < j0 → i1 + i0 < j + j0, i1 -1 < j, i0 < j0 ⊦ i1 -1 + i0 < j + j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i0 < j0, i1 < j + 1, ∀ j, i0, j0. i1 < j ∧ i0 < j0 → i1 + i0 < j + j0 ⊦ i0 + i1 < (j + j0) + 1

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1;%add-pred: ⊦ i + j -1 = (i + j)-1;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%ls-pred-04: ⊦ i -1 < j ↔ i < j +1; from specification int-basic1