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Proof step 52 from proof path-equiv-reach (previous step)

# y = 1 + n0, path(a0 ' + y), ∀ a, b. reach(a, b, n0) ↔ (∃ x. path(x) ∧ # x = n0 + 1 ∧ x .first = a ∧ x .last = b) ⊦ y = [], ∃ c. reach(a0, c, n0) ∧ y .last ∈ succ(c)

User applied rule all left

And obtained the following new goals

Proof step 51:
# y = 1 + n0, path(a0 ' + y), reach(a0, y .butlast .last, n0) ↔ (∃ x. path(x) ∧ # x = n0 + 1 ∧ x .first = a0 ∧ x .last = y .butlast .last) ⊦ y = [], ∃ c. reach(a0, c, n0) ∧ y .last ∈ succ(c)

Rule argument

[EXRARG : pos: left 3, [extquanttermlist: (a0, y .butlast .last) abort:F , discard:T , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (a0, y .butlast .last)]]