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Proof step 5 from proof reach-trans (previous step)

c ∈ succ(c0), reach(b, c0, m0), reach(a, b, n), ∀ a, b, n, c. reach(a, b, n) ∧ reach(b, c, m0) → reach(a, c, m0 + n) ⊦ succ(c0) = ∅, # succ(c0) = 0, ∃ c0. reach(a, c0, m0 + n) ∧ c ∈ succ(c0)

User applied rule all left

And obtained the following new goals

Proof step 4:
c ∈ succ(c0), reach(b, c0, m0), reach(a, b, n), reach(a, b, n) ∧ reach(b, c0, m0) → reach(a, c0, m0 + n) ⊦ succ(c0) = ∅, # succ(c0) = 0, ∃ c0. reach(a, c0, m0 + n) ∧ c ∈ succ(c0)

Rule argument

[EXRARG : pos: left 4, [extquanttermlist: (a, b, n, c0) abort:F , discard:T , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (a, b, n, c0)]]