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Proof step 5 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 6:
i +1 < j + i1, j < j0, j0 + i1 < j + i1, i < j + i1, ∀ i, j, j0. i < j + i1 ∧ j < j0 ∧ j0 + i1 < j + i1 → i < j0 + i1 ⊦ 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;%lf-04: ⊦ ¬ j < i ∧ j < i0 → i < i0; (forward) from specification int-basic1%lf: ⊦ i +1 < j → i < j; (forward) from specification int-basic1%ls-pred-01: ⊦ i -1 < j -1 ↔ i < j; from specification int-basic1%ls-pred: ⊦ i < j -1 ↔ i +1 < j; from specification int-basic1