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Proof step 3 from proof reach-one (conclusion)

⊦ reach(a, b, 1) ↔ b ∈ succ(a)

User applied rule insert lemma

And obtained the following new goals

Proof step 1:
⊦ reach(a, b, n + 1) ↔ (∃ c. reach(a, c, n) ∧ b ∈ succ(c))
Proof step 2:
⊦ reach(a, b, 0) ↔ a = b

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: reach-succ antecedent: succedent: reach(a, b, n + 1) ↔ (∃ c. c ∈ succ(a) ∧ reach(c, b, n)) substlist: [a, b, n] → [a, b, 0] add preconditions?: F is rewrite lemma?: F current spec?: T rotate lemma?: F replace all?: F paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (a, b, 0)]]