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Proof step 3 from proof ex-ls-add (previous step)

∀ j. i0 < j → (∃ i1. j = i0 + i1 ∧ 0 < i1), i0 +1 < j ⊦ ∃ i1. j = i0 +1 + i1 ∧ 0 < i1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i0 + 1 < j, ∀ j. i0 < j → (∃ i1. j = i0 + i1 ∧ 0 < i1) ⊦ ∃ i. j = (i0 + 1) + i ∧ 0 < i

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1;%add-a: ⊦ i + j + i0 = (i + j) + i0;