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Proof step 31 from proof BFS-returns-fail-implies-not-reachable (previous step)

dest ∈ S0, reachS(source, S0, n0), reachS(source, S, n0 + 1), ∀ n. ¬ reach(source, dest, n), ∀ b. b ∈ S1 ↔ reach<(source, b, n0 + 1) ∨ (∃ a. b ∈ succ(a) ∧ reachS(source, ∅ ++ a, n0) ∧ ¬ a ∈ S0), ∀ b. reachS(source, ∅ ++ b, n0 + 1) → b ∈ S ∨ (∃ a. a ∈ S0 ∧ b ∈ succ(a) ∧ ¬ b ∈ S1) ⊦ S0 = ∅, # S0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ reachS(a, S, n) ↔ (∀ b. b ∈ S → reach(a, b, n) ∧ ¬ reach<(a, b, n))
Proof step 30:
dest ∈ S0, ∀ b. b ∈ S0 → reach(source, b, n0) ∧ ¬ reach<(source, b, n0), reachS(source, S, n0 + 1), ∀ n. ¬ reach(source, dest, n), ∀ b. b ∈ S1 ↔ reach<(source, b, n0 + 1) ∨ (∃ a. b ∈ succ(a) ∧ reachS(source, ∅ ++ a, n0) ∧ ¬ a ∈ S0), ∀ b. reachS(source, ∅ ++ b, n0 + 1) → b ∈ S ∨ (∃ a. a ∈ S0 ∧ b ∈ succ(a) ∧ ¬ b ∈ S1) ⊦ S0 = ∅, # S0 = 0

Rule argument

[x0lemmaarg: spec-name: graph instance-name: lemma-name: reachset-def antecedent: succedent: reachS(a, S, n) ↔ (∀ b. b ∈ S → reach(a, b, n) ∧ ¬ reach<(a, b, n)) substlist: [n, S, a] → [n0, S0, source] add preconditions?: F is rewrite lemma?: T current spec?: F rotate lemma?: F replace all?: T paths: ((0, 0, 1)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (n0, S0, source)]]