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Proof step 6 from proof INV-step (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]