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Proof step 27 from proof INV-nextlevel (previous step)

reachS(source, N, n + 1) ⊦ b ∈ V ↔ reach<(source, b, 2 + n) ∨ (∃ a. b ∈ succ(a) ∧ (∃ c. reach(source, c, n) ∧ a ∈ succ(c)) ∧ ¬ reach<(source, a, n + 1) ∧ ¬ a ∈ N), ∃ b1. ¬ ((∃ c. reach(source, c, n) ∧ b1 ∈ succ(c)) ∧ ¬ reach<(source, b1, n + 1) → b1 ∈ N), ¬ (b ∈ V ↔ reach<(source, b, n + 1) ∨ (∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, 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 26:
b ∈ V ↔ reach<(source, b, n + 1) ∨ (∃ a. b ∈ succ(a) ∧ reach(source, a, n) ∧ ¬ reach<(source, a, n)), reachS(source, N, n + 1) ⊦ b ∈ V ↔ reach<(source, b, 2 + n) ∨ (∃ a. b ∈ succ(a) ∧ (∃ c. reach(source, c, n) ∧ a ∈ succ(c)) ∧ ¬ reach<(source, a, n + 1) ∧ ¬ a ∈ 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)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]