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Proof step 28 from proof BFS-returns-fail-implies-not-reachable (previous step)

dest ∈ S0, reachS(source, S, n0 + 1), reach(source, dest, n0), ∀ 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, reach<(source, dest, n0)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 4, [QUANTTERMLIST : (n0)]]

Additional information

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