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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

+1(s(x), y) → +1(x, y)
+1(s(x), y) → +1(x, s(y))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

+1(s(x), y) → +1(x, y)
+1(s(x), y) → +1(x, s(y))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


+1(s(x), y) → +1(x, y)
+1(s(x), y) → +1(x, s(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x1, x2)
s(x1)  =  s(x1)

Recursive path order with status [2].
Precedence:
+^12 > s1

Status:
s1: multiset
+^12: [1,2]

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.