YES
(ignored inputs)COMMENT translated from Cops 199
*** Computating Strongly Quasi-Reducible Parts ***
TRS:
   [ +(0,?x) -> ?x,
     +(+(?x,?y),?z) -> +(?x,+(?y,?z)),
     +(?x,?y) -> +(?y,?x) ]
Constructors: {0}
Defined function symbols: {+}
Constructor subsystem:
   [ ]
Rule part & Conj Part:
   [ +(0,?x) -> ?x ]
   [ +(+(?x,?y),?z) -> +(?x,+(?y,?z)),
     +(?x,?y) -> +(?y,?x) ]
*** Ground Confluence Check by Rewriting Induction ***
Sort: {Elem}
Signature:
   [ + : Elem,Elem -> Elem,
     0 : Elem ]
Rule Part:
   [ +(0,?x) -> ?x ]
Conjecture Part:
   [ +(+(?x,?y),?z) = +(?x,+(?y,?z)),
     +(?x,?y) = +(?y,?x) ]
Precedence (by weight): {(+,0),(0,1)}
Rule part is confluent.
R0 is ground confluent.
Check conj part consists of inductive theorems of R0.
Rules:
   [ +(0,?x) -> ?x ]
Conjectures:
   [ +(+(?x,?y),?z) = +(?x,+(?y,?z)),
     +(?x,?y) = +(?y,?x) ]
STEP 0
ES:
   [ +(+(?x,?y),?z) = +(?x,+(?y,?z)),
     +(?x,?y) = +(?y,?x) ]
HS:
   [ ]
ES0:
   [ +(+(?x,?y),?z) = +(?x,+(?y,?z)),
     +(?x,?y) = +(?y,?x) ]
HS0:
   [ ]
ES1:
   [ +(+(?x,?y),?z) = +(?x,+(?y,?z)),
     +(?x,?y) = +(?y,?x) ]
HS1:
   [ ]
Expand +(?x,?y) = +(?y,?x)
   [ ?x_1 = +(?x_1,0) ]
ES2:
   [ ?x_1 = +(?x_1,0),
     +(+(?x,?y),?z) = +(?x,+(?y,?z)) ]
HS2:
   [ +(?x,?y) -> +(?y,?x) ]
STEP 1
ES:
   [ ?x_1 = +(?x_1,0),
     +(+(?x,?y),?z) = +(?x,+(?y,?z)) ]
HS:
   [ +(?x,?y) -> +(?y,?x) ]
ES0:
   [ ?x_1 = +(?x_1,0),
     +(+(?x,?y),?z) = +(?x,+(?y,?z)) ]
HS0:
   [ +(?x,?y) -> +(?y,?x) ]
ES1:
   [ ?x_1 = +(?x_1,0),
     +(+(?x,?y),?z) = +(?x,+(?y,?z)) ]
HS1:
   [ +(?x,?y) -> +(?y,?x) ]
Expand +(?x_1,0) = ?x_1
   [ 0 = 0 ]
ES2:
   [ 0 = 0,
     +(+(?x,?y),?z) = +(?x,+(?y,?z)) ]
HS2:
   [ +(?x_1,0) -> ?x_1,
     +(?x,?y) -> +(?y,?x) ]
STEP 2
ES:
   [ 0 = 0,
     +(+(?x,?y),?z) = +(?x,+(?y,?z)) ]
HS:
   [ +(?x_1,0) -> ?x_1,
     +(?x,?y) -> +(?y,?x) ]
ES0:
   [ 0 = 0,
     +(+(?x,?y),?z) = +(?x,+(?y,?z)) ]
HS0:
   [ +(?x_1,0) -> ?x_1,
     +(?x,?y) -> +(?y,?x) ]
ES1:
   [ +(+(?x,?y),?z) = +(?x,+(?y,?z)) ]
HS1:
   [ +(?x_1,0) -> ?x_1,
     +(?x,?y) -> +(?y,?x) ]
Expand +(+(?x,?y),?z) = +(?x,+(?y,?z))
   [ +(?x_1,?z) = +(0,+(?x_1,?z)) ]
ES2:
   [ +(?x_1,?z) = +(0,+(?x_1,?z)) ]
HS2:
   [ +(+(?x,?y),?z) -> +(?x,+(?y,?z)),
     +(?x_1,0) -> ?x_1,
     +(?x,?y) -> +(?y,?x) ]
STEP 3
ES:
   [ +(?x_1,?z) = +(0,+(?x_1,?z)) ]
HS:
   [ +(+(?x,?y),?z) -> +(?x,+(?y,?z)),
     +(?x_1,0) -> ?x_1,
     +(?x,?y) -> +(?y,?x) ]
ES0:
   [ +(?x_1,?z) = +(?x_1,?z) ]
HS0:
   [ +(+(?x,?y),?z) -> +(?x,+(?y,?z)),
     +(?x_1,0) -> ?x_1,
     +(?x,?y) -> +(?y,?x) ]
ES1:
   [ ]
HS1:
   [ +(+(?x,?y),?z) -> +(?x,+(?y,?z)),
     +(?x_1,0) -> ?x_1,
     +(?x,?y) -> +(?y,?x) ]
Conj part consisits of inductive theorems of R0.
examples/fromCops/cr/199.trs: Success(GCR)
(5 msec.)