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Proof step 12 from proof INV-step (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 11:
v ∈ C, reachS(source, C, n), reachS(source, N, n + 1) ⊦ b ∈ V ∪ succ(v) ↔ reach<(source, b, n + 1) ∨ (∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n) ∧ (a = v ∨ ¬ a ∈ C)), # C = 0, C = ∅, ¬ (b ∈ V ↔ reach<(source, b, n + 1) ∨ (∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n) ∧ ¬ a ∈ C)), ∃ 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))

Rule argument

[EXRARG : pos: right 4, [extquanttermlist: (b)abort:F, discard:F, computed:T]]

Additional information

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