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Proof step 31 from proof INV-step (previous step)

v ∈ C, reachS(source, C, n), reachS(source, N, n + 1), b0 ∈ succ(c), reach(source, c, n) ⊦ succ(c) = ∅, # C = 0, C = ∅, # succ(c) = 0, b0 ∈ succ(v) \ V, b0 ∈ N, reach<(source, b0, n + 1), ∃ a. a ≠ v ∧ a ∈ C ∧ b0 ∈ succ(a) ∧ ¬ b0 ∈ V ∪ succ(v), ∃ b1. ¬ (b1 ∈ V ↔ reach<(source, b1, n + 1) ∨ (∃ a. b1 ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n) ∧ ¬ a ∈ C)), ¬ ((∃ c. reach(source, c, n) ∧ b0 ∈ succ(c)) ∧ ¬ reach<(source, b0, n + 1) → b0 ∈ N ∨ (∃ a. a ∈ C ∧ b0 ∈ succ(a) ∧ ¬ b0 ∈ V)), c ≠ v ∧ c ∈ C ∧ ¬ b0 ∈ V ∪ succ(v)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
v ∈ C, reachS(source, C, n), reachS(source, N, n + 1), b0 ∈ succ(c), reach(source, c, n), (∃ c. reach(source, c, n) ∧ b0 ∈ succ(c)) → a ∈ C ∧ b0 ∈ succ(a) ∧ ¬ b0 ∈ V ⊦ succ(c) = ∅, # C = 0, C = ∅, # succ(c) = 0, b0 ∈ succ(v) \ V, b0 ∈ N, reach<(source, b0, n + 1), ∃ a. a ≠ v ∧ a ∈ C ∧ b0 ∈ succ(a) ∧ ¬ b0 ∈ V ∪ succ(v), ∃ b1. ¬ (b1 ∈ V ↔ reach<(source, b1, n + 1) ∨ (∃ a. b1 ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n) ∧ ¬ a ∈ C)), c ≠ v ∧ c ∈ C ∧ ¬ b0 ∈ V ∪ succ(v)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]