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Proof step 47 from proof path-equiv-reach (previous step)

path(a0 ' + y0 + a '), ∀ x. ¬ path(x) ∨ # x ≠ # y0 + 1 ∨ x .first ≠ a0 ∨ x .last ≠ y0 .last ⊦ reach(a0, y0 .last, # y0), ∃ c. reach(a0, c, # y0) ∧ a ∈ succ(c)

User applied rule all left

And obtained the following new goals

Proof step 46:
path(a0 ' + y0 + a '), ¬ path(a0 + y0) ∨ # (a0 + y0) ≠ # y0 + 1 ∨ (a0 + y0) .first ≠ a0 ∨ (a0 + y0) .last ≠ y0 .last ⊦ reach(a0, y0 .last, # y0), ∃ c. reach(a0, c, # y0) ∧ a ∈ succ(c)

Rule argument

[EXRARG : pos: left 2, [extquanttermlist: (a0 + y0)abort:F, discard:T, computed:F]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (a0 + y0)]]