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Proof step 1 from proof three (conclusion)

⊦ i→n(1 + 1 + 1) = 2 + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ 0 < i → i→n(i) = i→n(i - 1) + 1

Rule argument

spec: <toplevel> name: Int2nat-pos seq: ⊦ 0 < i → i→n(i) = i→n(i - 1) + 1 subst: [i] → [1 + 1 + 1] flags: <all paths>

Additional information

Used terms: 1 + 1 + 1