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Proof step 18 from proof path-equiv-reach (previous step)

path(a ' + y), b ∈ succ(y .last), reach(a, y .last, # y) ⊦ y = [], succ(y .last) = ∅, # succ(y .last) = 0, ∃ x0. path(x0) ∧ # x0 = 2 + # y ∧ x0 .first = a ∧ x0 .last = b

User applied rule exists right

And obtained the following new goals

Proof step 17:
path(a ' + y), b ∈ succ(y .last), reach(a, y .last, # y) ⊦ y = [], succ(y .last) = ∅, # succ(y .last) = 0, path(a + (y + b)) ∧ # (a + (y + b)) = 2 + # y ∧ (a + (y + b)) .first = a ∧ (a + (y + b)) .last = b

Rule argument

[EXRARG : pos: right 4, [extquanttermlist: (a + (y + b)) abort:F , discard:T , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (a + (y + b))]]