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Proof step 15 from proof reach-rec-rev (previous step)

reach(a, b, 2 + n0), ∀ a0, b0. (∃ c. reach(a0, c, n0) ∧ b0 ∈ succ(c)) → (∃ c. c ∈ succ(a0) ∧ reach(c, b0, n0)) ⊦ ∃ c. c ∈ succ(a) ∧ (∃ c0. reach(c, c0, n0) ∧ b ∈ succ(c0))

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ reach(a, b, n + 1) ↔ (∃ c. reach(a, c, n) ∧ b ∈ succ(c))
Proof step 14:
reach(a, b, (n0 + 1) + 1) ↔ (∃ c. reach(a, c, n0 + 1) ∧ b ∈ succ(c)), reach(a, b, 2 + n0), ∀ a0, b0. (∃ c. reach(a0, c, n0) ∧ b0 ∈ succ(c)) → (∃ c. c ∈ succ(a0) ∧ reach(c, b0, n0)) ⊦ ∃ c. c ∈ succ(a) ∧ (∃ c0. reach(c, c0, n0) ∧ b ∈ succ(c0))

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: reach-succ antecedent: succedent: reach(a, b, n + 1) ↔ (∃ c. reach(a, c, n) ∧ b ∈ succ(c)) substlist: [a, b, n] → [a, b, n0 + 1] add preconditions?: F is rewrite lemma?: F current spec?: T rotate lemma?: F replace all?: F paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (a, b, n0 + 1)]]