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Proof step 1 from proof ex-ls-add (conclusion)

⊦ i < j ↔ (∃ i0. j = i + i0 ∧ 0 < i0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i < j ⊦ ∃ i0. j = i + i0 ∧ 0 < i0

Rule argument

Additional information

Used simplifier rules: add-less: ⊦ 0 < i → i1 < i1 + i;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;