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Proof step 8 from proof ex-ls-add (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ j. i0 < j → (∃ i1. j = i0 + i1 ∧ 0 < i1), i0 = j ∨ i0 < j ⊦ ∃ i. j + 1 = i0 + i ∧ 0 < i

Rule argument

Additional information

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