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Proof step 7 from proof pls (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: j + 1, i0, j0

Additional information

Used simplifier rules: add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%lst-02: ⊦ i < j + i0 ∧ ¬ j0 < j → i < j0 + i0; Used terms: j + 1, i0, j0