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Proof step 10 from proof INV-step (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 7:
b ∈ V ∪ succ(v), b ∈ V ↔ reach<(source, b, n + 1) ∨ (∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n) ∧ ¬ a ∈ C), reachS(source, N, n + 1), reachS(source, C, n), v ∈ C ⊦ reach<(source, b, n + 1) ∨ (∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n) ∧ (a = v ∨ ¬ a ∈ C)), C = ∅, # C = 0, ∃ 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))
Proof step 9:
reach<(source, b, n + 1) ∨ (∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n) ∧ (a = v ∨ ¬ a ∈ C)), b ∈ V ↔ reach<(source, b, n + 1) ∨ (∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n) ∧ ¬ a ∈ C), reachS(source, N, n + 1), reachS(source, C, n), v ∈ C ⊦ b ∈ V ∪ succ(v), C = ∅, # C = 0, ∃ 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

[CASEDARG : pos: right 3, ()]

Additional information

[[CONCRETESIMPRULES : ()]]