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Proof step 18 from proof INV-step (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
reachS(source, C, n), v ∈ C, b ∈ succ(v), reach<(source, b, n + 1) ⊦ # succ(v) = 0, N ∪ (succ(v) \ V) = ∅, C = ∅, # C = 0, # (N ∪ (succ(v) \ V)) = 0, succ(v) = ∅, b ∈ N, b ∈ V, ∃ 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))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]