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Proof step 40 from proof INV-nextlevel (previous step)

reachS(source, N, n + 1), reach(source, b0, 2 + n), reach(source, c0, n), c ∈ succ(c0), b0 ∈ succ(c) ⊦ ¬ reach<(source, c, n + 1), # succ(c0) = 0, # succ(c) = 0, succ(c0) = ∅, succ(c) = ∅, reach<(source, b0, 2 + n), c ∈ N ∧ ¬ b0 ∈ V, ∃ a. a ∈ N ∧ b0 ∈ succ(a) ∧ ¬ b0 ∈ V, ∃ b1. ¬ ((∃ c. reach(source, c, n) ∧ b1 ∈ succ(c)) ∧ ¬ reach<(source, b1, n + 1) → b1 ∈ N), ∃ b1. ¬ (b1 ∈ V ↔ reach<(source, b1, n + 1) ∨ (∃ a. b1 ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
b0 ∈ succ(c), c ∈ succ(c0), reach(source, c0, n), reach(source, b0, 2 + n), reachS(source, N, n + 1), reach<(source, c, n + 1) ⊦ succ(c) = ∅, succ(c0) = ∅, # succ(c) = 0, # succ(c0) = 0, reach<(source, b0, 2 + n), ∃ b1. ¬ (b1 ∈ V ↔ reach<(source, b1, n + 1) ∨ (∃ a. b1 ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n))), ∃ b1. ¬ ((∃ c. reach(source, c, n) ∧ b1 ∈ succ(c)) ∧ ¬ reach<(source, b1, n + 1) → b1 ∈ N), ∃ a. a ∈ N ∧ b0 ∈ succ(a) ∧ ¬ b0 ∈ V, c ∈ N ∧ ¬ b0 ∈ V

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]