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

The TRS R consists of the following rules:

+(*(x, y), *(x, z)) → *(x, +(y, z))
+(+(x, y), 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
      ↳ EdgeDeletionProof

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

+1(+(x, y), z) → +1(y, z)
+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, z))

The TRS R consists of the following rules:

+(*(x, y), *(x, z)) → *(x, +(y, z))
+(+(x, y), 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 deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

+1(+(x, y), z) → +1(y, z)
+1(*(x, y), *(x, z)) → +1(y, z)
+1(*(x, y), +(*(x, z), u)) → +1(y, z)
+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), *(x, z)) → *(x, +(y, z))
+(+(x, y), 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 approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

+(*(x, y), *(x, z)) → *(x, +(y, z))
+(+(x, y), 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(y, z)
+1(*(x, y), +(*(x, z), u)) → +1(y, z)
+1(*(x, y), *(x, z)) → +1(y, z)
+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)  =  +1(x1)
+(x1, x2)  =  +(x1, x2)
*(x1, x2)  =  *(x2)
u  =  u

Recursive path order with status [2].
Precedence:
+2 > +^11
+2 > u
*1 > +^11
*1 > u

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

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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

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