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Proof step 24 from proof path-equiv-reach (previous step)

b ∈ succ(c), reach(a, c, n0), ∀ a, b. reach(a, b, n0) ↔ (∃ x. path(x) ∧ # x = n0 + 1 ∧ x .first = a ∧ x .last = b) ⊦ succ(c) = ∅, # succ(c) = 0, ∃ x. path(x) ∧ # x = 2 + n0 ∧ x .first = a ∧ x .last = b

User applied rule all left

And obtained the following new goals

Proof step 23:
b ∈ succ(c), reach(a, c, n0), reach(a, c, n0) ↔ (∃ x. path(x) ∧ # x = n0 + 1 ∧ x .first = a ∧ x .last = c) ⊦ succ(c) = ∅, # succ(c) = 0, ∃ x. path(x) ∧ # x = 2 + n0 ∧ x .first = a ∧ x .last = b

Rule argument

[EXRARG : pos: left 3, [extquanttermlist: (a, c)abort:F, discard:T, computed:T]]

Additional information

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