YES
*** Computating Strongly Quasi-Reducible Parts ***
TRS:
   [ +(0,0) -> 0,
     +(s(?x),?y) -> s(+(?x,?y)),
     +(?x,s(?y)) -> s(+(?y,?x)) ]
Constructors: {0,s}
Defined function symbols: {+}
Constructor subsystem:
   [ ]
Rule part & Conj Part:
   [ +(0,0) -> 0,
     +(s(?x),?y) -> s(+(?x,?y)),
     +(?x,s(?y)) -> s(+(?y,?x)) ]
   [ ]
*** Ground Confluence Check by Rewriting Induction ***
Sort: {Nat}
Signature:
   [ + : Nat,Nat -> Nat,
     s : Nat -> Nat,
     0 : Nat ]
Rule Part:
   [ +(0,0) -> 0,
     +(s(?x),?y) -> s(+(?x,?y)),
     +(?x,s(?y)) -> s(+(?y,?x)) ]
Conjecture Part:
   [ ]
Precedence (by weight): {(+,3),(0,1),(s,0)}
Rule part is not confluent.
Check ground confluence of Rule part (R0) by basic RI.
Rules:
   [ +(0,0) -> 0,
     +(s(?x),?y) -> s(+(?x,?y)),
     +(?x,s(?y)) -> s(+(?y,?x)) ]
Conjectures:
   [ s(s(+(?x,?y_1))) = s(s(+(?y_1,?x))) ]
STEP 0
ES:
   [ s(s(+(?x,?y_1))) = s(s(+(?y_1,?x))) ]
HS:
   [ ]
ES0:
   [ s(s(+(?x,?y_1))) = s(s(+(?y_1,?x))) ]
HS0:
   [ ]
ES1:
   [ s(s(+(?x,?y_1))) = s(s(+(?y_1,?x))) ]
HS1:
   [ ]
Expand s(s(+(?x,?y_1))) = s(s(+(?y_1,?x)))
   [ s(s(0)) = s(s(+(0,0))),
     s(s(s(+(?x_2,?y_2)))) = s(s(+(?y_2,s(?x_2)))),
     s(s(s(+(?y_3,?x_3)))) = s(s(+(s(?y_3),?x_3))) ]
ES2:
   [ s(s(0)) = s(s(+(0,0))),
     s(s(s(+(?x_2,?y_2)))) = s(s(+(?y_2,s(?x_2)))),
     s(s(s(+(?y_3,?x_3)))) = s(s(+(s(?y_3),?x_3))) ]
HS2:
   [ s(s(+(?x,?y_1))) -> s(s(+(?y_1,?x))) ]
STEP 1
ES:
   [ s(s(0)) = s(s(+(0,0))),
     s(s(s(+(?x_2,?y_2)))) = s(s(+(?y_2,s(?x_2)))),
     s(s(s(+(?y_3,?x_3)))) = s(s(+(s(?y_3),?x_3))) ]
HS:
   [ s(s(+(?x,?y_1))) -> s(s(+(?y_1,?x))) ]
ES0:
   [ s(s(0)) = s(s(0)),
     s(s(s(+(?x_2,?y_2)))) = s(s(s(+(?x_2,?y_2)))),
     s(s(s(+(?y_3,?x_3)))) = s(s(s(+(?y_3,?x_3)))) ]
HS0:
   [ s(s(+(?x,?y_1))) -> s(s(+(?y_1,?x))) ]
ES1:
   [ ]
HS1:
   [ s(s(+(?x,?y_1))) -> s(s(+(?y_1,?x))) ]
R0 is ground confluent.
Conjucture part is empty.
examples/additions/plus1.trs: Success(GCR)
(8 msec.)