MAYBE
(ignored inputs)COMMENT translated from Cops 128
*** Computating Strongly Quasi-Reducible Parts ***
TRS:
   [ +(0,?y) -> ?y,
     +(?x,s(?y)) -> s(+(?x,?y)),
     +(?x,?y) -> +(?y,?x) ]
Constructors: {+,0,s}
Defined function symbols: {}
Constructor subsystem:
   [ +(0,?y) -> ?y,
     +(?x,s(?y)) -> s(+(?x,?y)) ]
Rule part & Conj Part:
   [ +(0,?y) -> ?y,
     +(?x,s(?y)) -> s(+(?x,?y)) ]
   [ +(?x,?y) -> +(?y,?x) ]
*** Ground Confluence Check by Rewriting Induction ***
Sort: {Nat}
Signature:
   [ + : Nat,Nat -> Nat,
     0 : Nat,
     s : Nat -> Nat ]
Rule Part:
   [ +(0,?y) -> ?y,
     +(?x,s(?y)) -> s(+(?x,?y)) ]
Conjecture Part:
   [ +(?x,?y) = +(?y,?x) ]
Precedence (by weight): {(+,3),(0,2),(s,0)}
Rule part is confluent.
R0 is ground confluent.
Check conj part consists of inductive theorems of R0.
Rules:
   [ +(0,?y) -> ?y,
     +(?x,s(?y)) -> s(+(?x,?y)) ]
Conjectures:
   [ +(?x,?y) = +(?y,?x) ]
STEP 0
ES:
   [ +(?x,?y) = +(?y,?x) ]
HS:
   [ ]
ES0:
   [ +(?x,?y) = +(?y,?x) ]
HS0:
   [ ]
ES1:
   [ +(?x,?y) = +(?y,?x) ]
HS1:
   [ ]
No equation to expand
Failed to prove conj part consists of inductive theorems of R0.
examples/fromCops/cr/128.trs: Failure(unknown)
(10 msec.)