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Proof step 1 from proof get-from-inf (conclusion) (back to tree)

⊦ ((nf from n))(m) = nf(m + n)

System applied rule normalize

And obtained the following new goals

Proof step 2:
((nf from n))(m) ≠ nf(m + n) ⊦

Rule argument

Additional information

Proof step 2 from proof get-from-inf (previous step) (back to tree)

((nf from n))(m) ≠ nf(m + n) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ nf from n = (λ m. nf(n + m))

Rule argument

spec: <toplevel>, name:from-nf, seq: ⊦ nf from n = (λ m. nf(n + m)), subst:[n, nf] → [n, nf], <all paths>

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic Used terms: n, nf

Proof step 3 from proof get-from-inf (previous step) (back to tree)

⊦ nf from n = (λ m. nf(n + m))

Lemma from-nf

Proof Tree for get-from-inf (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.