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Proof step 3 from proof INV-step (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
b ∈ succ(v), reachS(source, N, n + 1), reachS(source, C, n), v ∈ C ⊦ succ(v) = ∅, # (V ∪ succ(v)) = 0, C = ∅, # C = 0, V ∪ succ(v) = ∅, # succ(v) = 0, reach<(source, b, n + 1), b ∈ V, ∃ 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)), ∃ a. b ∈ 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 : ⊦ s ∪ s0 = s0 ∪ s], [CSIMPSEQ : ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1], [CSIMPSEQ : ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2])]]