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Proof step 6 from proof ex-ls-add (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ i < j ↔ i +1 = j ∨ i +1 < j
Proof step 8:
∀ j. i0 < j → (∃ i1. j = i0 + i1 ∧ 0 < i1), i0 +1 = j + 1 ∨ i0 +1 < j + 1 ⊦ ∃ i. j + 1 = i + i0 ∧ 0 < i

Rule argument

spec: int-basic1 name: lsrec seq: ⊦ i < j ↔ i +1 = j ∨ i +1 < j subst: [j, i] → [j + 1, i0] flags: <all paths>

Additional information

Used terms: j + 1, i0