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Proof step 4 from proof INV-step (previous step)

b ∈ V ∪ succ(v), reachS(source, N, n + 1), reachS(source, C, n), v ∈ C, reach(source, a, n), b ∈ succ(a), b ∈ V ⊦ succ(a) = ∅, V = ∅, # (V ∪ succ(v)) = 0, C = ∅, # C = 0, V ∪ succ(v) = ∅, # succ(a) = 0, # V = 0, reach<(source, a, n), a ∈ C, reach<(source, b, n + 1), ∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n) ∧ (a = v ∨ ¬ a ∈ C), ∃ b1. ¬ ((∃ c. reach(source, c, n) ∧ b1 ∈ succ(c)) ∧ ¬ reach<(source, b1, n + 1) → b1 ∈ N ∨ (∃ a. a ∈ C ∧ b1 ∈ succ(a) ∧ ¬ b1 ∈ V)), ∃ b1. ¬ (b1 ∈ V ↔ reach<(source, b1, n + 1) ∨ (∃ a. b1 ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n) ∧ ¬ a ∈ C))

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

[EXRARG : pos: right 12, [QUANTTERMLIST : (a)]]

Additional information

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