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Proof step 6 from proof lst-02 (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: i + 1, j, j0

Additional information

Used simplifier rules: add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; Used terms: i + 1, j, j0