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Proof step 1 from proof nat2int-03 (conclusion)

⊦ n→i(n + 1) = n→i(n) + 1

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ n→i(n +1) = n→i(n) +1

Rule argument

spec: <toplevel> name: Nat2int-succ seq: ⊦ n→i(n +1) = n→i(n) +1 subst: [n] → [n]

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ i +1 = i + 1; (s) Used terms: n