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:

+(x, +(y, z)) → +(+(x, y), z)
+(*(x, y), +(x, z)) → *(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+(x, +(y, z)) → +(+(x, y), z)
+(*(x, y), +(x, z)) → *(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)

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(*(x, y), +(*(x, z), u)) → +1(y, z)
+1(*(x, y), +(x, z)) → +1(y, z)
+1(*(x, y), +(*(x, z), u)) → +1(*(x, +(y, z)), u)
+1(x, +(y, z)) → +1(x, y)
+1(x, +(y, z)) → +1(+(x, y), z)

The TRS R consists of the following rules:

+(x, +(y, z)) → +(+(x, y), z)
+(*(x, y), +(x, z)) → *(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)

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(*(x, y), +(*(x, z), u)) → +1(y, z)
+1(*(x, y), +(x, z)) → +1(y, z)
+1(*(x, y), +(*(x, z), u)) → +1(*(x, +(y, z)), u)
+1(x, +(y, z)) → +1(x, y)
+1(x, +(y, z)) → +1(+(x, y), z)

The TRS R consists of the following rules:

+(x, +(y, z)) → +(+(x, y), z)
+(*(x, y), +(x, z)) → *(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)

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(*(x, y), +(*(x, z), u)) → +1(y, z)
+1(*(x, y), +(x, z)) → +1(y, z)
The remaining pairs can at least be oriented weakly.

+1(*(x, y), +(*(x, z), u)) → +1(*(x, +(y, z)), u)
+1(x, +(y, z)) → +1(x, y)
+1(x, +(y, z)) → +1(+(x, y), z)
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x1)
*(x1, x2)  =  *(x1, x2)
+(x1, x2)  =  x1

Lexicographic path order with status [19].
Precedence:
trivial

Status:
+^11: [1]
*2: [2,1]

The following usable rules [14] were oriented:

+(x, +(y, z)) → +(+(x, y), z)
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)
+(*(x, y), +(x, z)) → *(x, +(y, z))



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

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

+1(x, +(y, z)) → +1(x, y)
+1(*(x, y), +(*(x, z), u)) → +1(*(x, +(y, z)), u)
+1(x, +(y, z)) → +1(+(x, y), z)

The TRS R consists of the following rules:

+(x, +(y, z)) → +(+(x, y), z)
+(*(x, y), +(x, z)) → *(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)

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

Lexicographic path order with status [19].
Precedence:
trivial

Status:
+2: [1,2]
*2: [1,2]

The following usable rules [14] were oriented: none



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

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

+(x, +(y, z)) → +(+(x, y), z)
+(*(x, y), +(x, z)) → *(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)

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.