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Proof step 5 from proof reachless-succ-edge (previous step)

reach(source, a, n0), b ∈ succ(a) ⊦ succ(a) = ∅, # succ(a) = 0, ∃ m. m < n0 + 2 ∧ reach(source, b, m)

User applied rule exists right

And obtained the following new goals

Proof step 4:
reach(source, a, n0), b ∈ succ(a) ⊦ succ(a) = ∅, # succ(a) = 0, n0 + 1 < n0 + 2 ∧ reach(source, b, n0 + 1)

Rule argument

[EXRARG : pos: right 3, [extquanttermlist: (n0 + 1)abort:F, discard:T, computed:F]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (n0 + 1)]]