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))
+(f(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+(+(x, y), z) → +(x, +(y, z))
+(f(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

+(+(x, y), z) → +(x, +(y, z))
+(f(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)

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

The TRS R consists of the following rules:

+(+(x, y), z) → +(x, +(y, z))
+(f(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


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

+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1)) = 1/4 + x_1   
POL(+1(x1, x2)) = (1/2)x_1 + (1/2)x_2   
POL(+(x1, x2)) = x_1 + x_2   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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

The TRS R consists of the following rules:

+(+(x, y), z) → +(x, +(y, z))
+(f(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1)) = 15/4 + x_1   
POL(+1(x1, x2)) = x_1   
POL(+(x1, x2)) = 4 + (4)x_1 + x_2   
The value of delta used in the strict ordering is 4.
The following usable rules [17] 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))
+(f(x), f(y)) → f(+(x, y))
+(f(x), +(f(y), z)) → +(f(+(x, y)), z)

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.