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Proof step 11 from proof INV-nextlevel (previous step)

b ∈ succ(c), reach(source, c, n), reachS(source, N, n + 1) ⊦ succ(c) = ∅, # succ(c) = 0, b ∈ V, ∃ m0. ¬ n < m0 ∧ reach(source, b, m0), ∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n), ∃ b1. ¬ ((∃ c. reach(source, c, n) ∧ b1 ∈ succ(c)) ∧ ¬ reach<(source, b1, n + 1) → b1 ∈ N), ∃ b1. ¬ (b1 ∈ V ↔ reach<(source, b1, n + 1) ∨ (∃ a. b1 ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n)))

User applied rule exists right

And obtained the following new goals

Proof step 10:
b ∈ succ(c), reach(source, c, n), reachS(source, N, n + 1) ⊦ succ(c) = ∅, # succ(c) = 0, b ∈ V, ∃ m0. ¬ n < m0 ∧ reach(source, b, m0), b ∈ succ(c) ∧ reach(source, c, n) ∧ ¬ reach<(source, c, n), ∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n), ∃ b1. ¬ ((∃ c. reach(source, c, n) ∧ b1 ∈ succ(c)) ∧ ¬ reach<(source, b1, n + 1) → b1 ∈ N), ∃ b1. ¬ (b1 ∈ V ↔ reach<(source, b1, n + 1) ∨ (∃ a. b1 ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n)))

Rule argument

[EXRARG : pos: right 5, [extquanttermlist: (c)abort:F, discard:F, computed:T]]

Additional information

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