YES
*** Computating Strongly Quasi-Reducible Parts ***
TRS:
   [ +(0,?y) -> ?y,
     +(s(0),?y) -> s(+(0,?y)),
     +(?x,?y) -> +(?y,?x),
     s(s(?x)) -> ?x ]
Constructors: {0,s}
Defined function symbols: {+}
Constructor subsystem:
   [ s(s(?x)) -> ?x ]
Rule part & Conj Part:
   [ s(s(?x)) -> ?x,
     +(0,?y) -> ?y,
     +(s(0),?y) -> s(+(0,?y)) ]
   [ +(?x,?y) -> +(?y,?x) ]
*** Ground Confluence Check by Rewriting Induction ***
Sort: {Nat}
Signature:
   [ + : Nat,Nat -> Nat,
     s : Nat -> Nat,
     0 : Nat ]
Rule Part:
   [ s(s(?x)) -> ?x,
     +(0,?y) -> ?y,
     +(s(0),?y) -> s(+(0,?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:
   [ s(s(?x)) -> ?x,
     +(0,?y) -> ?y,
     +(s(0),?y) -> s(+(0,?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:
   [ ]
Expand +(?x,?y) = +(?y,?x)
   [ ?y_2 = +(?y_2,0),
     s(+(0,?y_3)) = +(?y_3,s(0)),
     +(?x_5,?x_4) = +(?x_4,s(s(?x_5))) ]
ES2:
   [ ?y_2 = +(?y_2,0),
     s(?y_3) = +(?y_3,s(0)),
     +(?x_5,?x_4) = +(?x_4,s(s(?x_5))) ]
HS2:
   [ +(?x,?y) -> +(?y,?x) ]
STEP 1
ES:
   [ ?y_2 = +(?y_2,0),
     s(?y_3) = +(?y_3,s(0)),
     +(?x_5,?x_4) = +(?x_4,s(s(?x_5))) ]
HS:
   [ +(?x,?y) -> +(?y,?x) ]
ES0:
   [ ?y_2 = +(?y_2,0),
     s(?y_3) = +(?y_3,s(0)),
     +(?x_5,?x_4) = +(?x_4,?x_5) ]
HS0:
   [ +(?x,?y) -> +(?y,?x) ]
ES1:
   [ ?y_2 = +(?y_2,0),
     s(?y_3) = +(?y_3,s(0)) ]
HS1:
   [ +(?x,?y) -> +(?y,?x) ]
Expand +(?y_2,0) = ?y_2
   [ 0 = 0,
     s(+(0,0)) = s(0),
     +(?x_5,0) = s(s(?x_5)) ]
ES2:
   [ 0 = 0,
     s(0) = s(0),
     +(?x_5,0) = s(s(?x_5)),
     s(?y_3) = +(?y_3,s(0)) ]
HS2:
   [ +(?y_2,0) -> ?y_2,
     +(?x,?y) -> +(?y,?x) ]
STEP 2
ES:
   [ 0 = 0,
     s(0) = s(0),
     +(?x_5,0) = s(s(?x_5)),
     s(?y_3) = +(?y_3,s(0)) ]
HS:
   [ +(?y_2,0) -> ?y_2,
     +(?x,?y) -> +(?y,?x) ]
ES0:
   [ 0 = 0,
     s(0) = s(0),
     ?x_5 = ?x_5,
     s(?y_3) = +(?y_3,s(0)) ]
HS0:
   [ +(?y_2,0) -> ?y_2,
     +(?x,?y) -> +(?y,?x) ]
ES1:
   [ s(?y_3) = +(?y_3,s(0)) ]
HS1:
   [ +(?y_2,0) -> ?y_2,
     +(?x,?y) -> +(?y,?x) ]
Expand +(?y_3,s(0)) = s(?y_3)
   [ s(0) = s(0),
     s(+(0,s(0))) = s(s(0)),
     +(?x_5,s(0)) = s(s(s(?x_5))) ]
ES2:
   [ s(0) = s(0),
     0 = s(s(0)),
     +(?x_5,s(0)) = s(s(s(?x_5))) ]
HS2:
   [ +(?y_3,s(0)) -> s(?y_3),
     +(?y_2,0) -> ?y_2,
     +(?x,?y) -> +(?y,?x) ]
STEP 3
ES:
   [ s(0) = s(0),
     0 = s(s(0)),
     +(?x_5,s(0)) = s(s(s(?x_5))) ]
HS:
   [ +(?y_3,s(0)) -> s(?y_3),
     +(?y_2,0) -> ?y_2,
     +(?x,?y) -> +(?y,?x) ]
ES0:
   [ s(0) = s(0),
     0 = 0,
     s(?x_5) = s(?x_5) ]
HS0:
   [ +(?y_3,s(0)) -> s(?y_3),
     +(?y_2,0) -> ?y_2,
     +(?x,?y) -> +(?y,?x) ]
ES1:
   [ ]
HS1:
   [ +(?y_3,s(0)) -> s(?y_3),
     +(?y_2,0) -> ?y_2,
     +(?x,?y) -> +(?y,?x) ]
Conj part consisits of inductive theorems of R0.
examples/additions/plus11.trs: Success(GCR)
(9 msec.)