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Proof step 2 from proof ex-ls-add (previous step)

i < j ⊦ ∃ i0. j = i + i0 ∧ 0 < i0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ j. i0 < j → (∃ i1. j = i0 + i1 ∧ 0 < i1), i0 +1 < j ⊦ ∃ i1. j = i0 +1 + i1 ∧ 0 < i1
Proof step 5:
∀ j. i0 < j → (∃ i1. j = i0 + i1 ∧ 0 < i1), i0 -1 < j ⊦ ∃ i1. j = i0 -1 + i1 ∧ 0 < i1

Rule argument

indvar: i

Additional information

Used simplifier rules: add-zero-rev: ⊦ 0 + i = i;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%int generated by 0, +1, -1;;