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Proof step 6 from proof lst-01 (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: j0 + i1, j, j0

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1;%ls-01: ⊦ i + j < i + i0 ↔ j < i0;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: j0 + i1, j, j0