Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

Q restricted rewrite system:
The TRS R consists of the following rules:

+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))

Q is empty.

The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

+(s(x), s(y)) → s(+(s(x), +(y, 0)))
Used ordering:
Polynomial interpretation [25]:

POL(+(x1, x2)) = x1 + 2·x2   
POL(0) = 0   
POL(s(x1)) = 2 + x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

Q restricted rewrite system:
The TRS R consists of the following rules:

+(0, y) → y
+(s(x), 0) → s(x)

Q is empty.

The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

+(0, y) → y
+(s(x), 0) → s(x)

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

+(0, y) → y
+(s(x), 0) → s(x)
Used ordering:
Polynomial interpretation [25]:

POL(+(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(0) = 2   
POL(s(x1)) = 2 + 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RisEmptyProof

Q restricted rewrite system:
R is empty.
Q is empty.

The TRS R is empty. Hence, termination is trivially proven.