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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
j < j0, j0 + i1 < i, j0 + i1 < j + i1, ∀ i, j, j0. i < j + i1 ∧ j < j0 ∧ j0 + i1 < j + i1 → i < j0 + i1 ⊦ j + i1 < i

Rule argument

Additional information

Used simplifier rules: add-succ: ⊦ i + j +1 = (i + j)+1;%add-a: ⊦ i + j + i0 = (i + j) + i0;%lf-07: ⊦ ¬ i < i0 ∧ j < i0 → j < i; (forward) from specification int-basic1%ls-succ-02: ⊦ i +1 < j +1 ↔ i < j; from specification int-basic1%ls-succ: ⊦ i < j +1 ↔ ¬ j < i; from specification int-basic1