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Proof step 13 from proof reach-rec-rev (previous step)

b ∈ succ(c), c ∈ succ(c0), reach(a, c0, n0), reach(a, b, 2 + n0), ∀ a0, b0. (∃ c. reach(a0, c, n0) ∧ b0 ∈ succ(c)) → (∃ c. c ∈ succ(a0) ∧ reach(c, b0, n0)) ⊦ succ(c) = ∅, succ(c0) = ∅, # succ(c) = 0, # succ(c0) = 0, ∃ c. c ∈ succ(a) ∧ (∃ c0. reach(c, c0, n0) ∧ b ∈ succ(c0))

User applied rule all left

And obtained the following new goals

Proof step 12:
b ∈ succ(c), c ∈ succ(c0), reach(a, c0, n0), reach(a, b, 2 + n0), (∃ c0. reach(a, c0, n0) ∧ c ∈ succ(c0)) → (∃ c0. c0 ∈ succ(a) ∧ reach(c0, c, n0)) ⊦ succ(c) = ∅, succ(c0) = ∅, # succ(c) = 0, # succ(c0) = 0, ∃ c. c ∈ succ(a) ∧ (∃ c0. reach(c, c0, n0) ∧ b ∈ succ(c0))

Rule argument

[EXRARG : pos: left 5, [extquanttermlist: (a, c)abort:F, discard:T, computed:F]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (a, c)]]