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Proof step 30 from proof BFS-returns-fail-implies-not-reachable (previous step)

dest ∈ S0, ∀ b. b ∈ S0 → reach(source, b, n0) ∧ ¬ reach<(source, b, n0), reachS(source, S, n0 + 1), ∀ n. ¬ reach(source, dest, n), ∀ b. b ∈ S1 ↔ reach<(source, b, n0 + 1) ∨ (∃ a. b ∈ succ(a) ∧ reachS(source, ∅ ++ a, n0) ∧ ¬ a ∈ S0), ∀ b. reachS(source, ∅ ++ b, n0 + 1) → b ∈ S ∨ (∃ a. a ∈ S0 ∧ b ∈ succ(a) ∧ ¬ b ∈ S1) ⊦ S0 = ∅, # S0 = 0

User applied rule all left

And obtained the following new goals

Proof step 29:
dest ∈ S0, dest ∈ S0 → reach(source, dest, n0) ∧ ¬ reach<(source, dest, n0), reachS(source, S, n0 + 1), ∀ n. ¬ reach(source, dest, n), ∀ b. b ∈ S1 ↔ reach<(source, b, n0 + 1) ∨ (∃ a. b ∈ succ(a) ∧ reachS(source, ∅ ++ a, n0) ∧ ¬ a ∈ S0), ∀ b. reachS(source, ∅ ++ b, n0 + 1) → b ∈ S ∨ (∃ a. a ∈ S0 ∧ b ∈ succ(a) ∧ ¬ b ∈ S1) ⊦ S0 = ∅, # S0 = 0

Rule argument

[EXRARG : pos: left 2, [extquanttermlist: (dest)abort:F, discard:T, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (dest)]]