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Proof step 22 from proof INV-step (previous step)

v ∈ C, reachS(source, C, n), ∀ b. b ∈ N → (∃ c. reach(source, c, n) ∧ b ∈ succ(c)) ∧ ¬ reach<(source, b, n + 1) ⊦ # C = 0, C = ∅, ∀ b. b ∈ N ∪ (succ(v) \ V) → reach(source, b, n + 1) ∧ ¬ reach<(source, b, n + 1), ∃ 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 21:
reachS(source, C, n), v ∈ C, b ∈ N ∪ (succ(v) \ V), ∀ b. b ∈ N → (∃ c. reach(source, c, n) ∧ b ∈ succ(c)) ∧ ¬ reach<(source, b, n + 1) ⊦ N ∪ (succ(v) \ V) = ∅, C = ∅, # C = 0, # (N ∪ (succ(v) \ V)) = 0, (∃ c. reach(source, c, n) ∧ b ∈ succ(c)) ∧ ¬ reach<(source, b, n + 1), ∃ 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 : ([CSIMPFORWARD : ⊦ s ≠ ∅ → # s ≠ 0], [CSIMPFORWARD : ⊦ a ∈ s → s ≠ ∅], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ reach(a, b, n + 1) ↔ (∃ c. reach(a, c, n) ∧ b ∈ succ(c))])]]