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Proof step 5 from proof ex-ls-add (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i0 < j + 1, ∀ j. i0 < j → (∃ i1. j = i0 + i1 ∧ 0 < i1) ⊦ ∃ i. j + 1 = i + i0 ∧ 0 < i

Rule argument

Additional information

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