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Proof step 20 from proof INV-nextlevel (previous step)

reachS(source, N, n + 1), a ∈ succ(c), reach(source, c, n), b ∈ succ(a) ⊦ succ(c) = ∅, succ(a) = ∅, # succ(c) = 0, # succ(a) = 0, reach<(source, a, n + 1), a ∈ N, reach<(source, b, n + 1), b ∈ V, reach(source, a, 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)))

User applied rule exists right

And obtained the following new goals

Proof step 19:
reachS(source, N, n + 1), a ∈ succ(c), reach(source, c, n), b ∈ succ(a) ⊦ succ(c) = ∅, succ(a) = ∅, # succ(c) = 0, # succ(a) = 0, reach<(source, a, n + 1), a ∈ N, reach<(source, b, n + 1), b ∈ V, reach(source, a, n), ∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n), ¬ ((∃ c. reach(source, c, n) ∧ a ∈ succ(c)) ∧ ¬ reach<(source, a, n + 1) → 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 11, [extquanttermlist: (a)abort:F, discard:F, computed:T]]

Additional information

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