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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(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:

F(+(x, y), z) → F(y, z)
F(+(x, y), z) → F(x, z)
F(+(x, y), z) → +1(f(x, z), f(y, z))
+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
+1(a, b) → +1(b, a)
+1(a, +(b, z)) → +1(b, +(a, z))
+1(a, +(b, z)) → +1(a, z)

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

F(+(x, y), z) → F(y, z)
F(+(x, y), z) → F(x, z)
F(+(x, y), z) → +1(f(x, z), f(y, z))
+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
+1(a, b) → +1(b, a)
+1(a, +(b, z)) → +1(b, +(a, z))
+1(a, +(b, z)) → +1(a, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

+1(a, +(b, z)) → +1(a, z)

The TRS R consists of the following rules:

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

POL(a) = 0   
POL(b) = 1   
POL(+1(x1, x2)) = (1/4)x_2   
POL(+(x1, x2)) = (1/4)x_1 + (4)x_2   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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:

+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(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(a) = 0   
POL(b) = 0   
POL(+1(x1, x2)) = (4)x_1   
POL(+(x1, x2)) = 4 + (4)x_1 + (2)x_2   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(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.


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

POL(F(x1, x2)) = (2)x_1   
POL(+(x1, x2)) = 1 + x_1 + (4)x_2   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

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

+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(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.