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Proof step 51 from proof INVR-step (previous step)

v ∈ C, reach(v, c, m), dest ∈ succ(c), reachS(source, N, n + 1), reachS(source, C, n), reach(source, c0, m + n), dest ∈ succ(c0), reach(c1, dest, m), c1 ∈ succ(v), ∀ b. ¬ b ∈ N ∪ (succ(v) \ V) ∨ ¬ reach(b, dest, m) ∨ reach<(b, dest, m), ∀ a. a = v ∨ ¬ a ∈ C ∨ (∀ c1. ¬ reach(a, c1, m) ∨ ¬ dest ∈ succ(c1)) ∨ reach<(a, dest, m + 1) ⊦ succ(v) = ∅, # succ(c) = 0, # succ(c0) = 0, C = ∅, v = dest, # C = 0, succ(c) = ∅, succ(c0) = ∅, # succ(v) = 0, reach<(v, dest, m + 1), reach<(source, dest, (m + 1) + n), ¬ ((∃ c. reach(source, c, n) ∧ c1 ∈ succ(c)) ∧ ¬ reach<(source, c1, n + 1) → c1 ∈ N ∨ (∃ a. a ∈ C ∧ c1 ∈ succ(a) ∧ ¬ c1 ∈ V)), ∃ b0. ¬ ((∃ c. reach(source, c, n) ∧ b0 ∈ succ(c)) ∧ ¬ reach<(source, b0, n + 1) → b0 ∈ N ∨ (∃ a. a ∈ C ∧ b0 ∈ succ(a) ∧ ¬ b0 ∈ V)), ∃ b0. ¬ (b0 ∈ V ↔ reach<(source, b0, n + 1) ∨ (∃ a0. b0 ∈ succ(a0) ∧ reach(source, a0, n) ∧ ¬ reach<(source, a0, n) ∧ ¬ a0 ∈ C))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]