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:

minus(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(x)

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(x, minus(y)) → +1(minus(x), y)
+1(x, +(y, z)) → +1(x, y)
+1(x, minus(y)) → MINUS(+(minus(x), y))
+1(x, +(y, z)) → +1(+(x, y), z)
+1(x, minus(y)) → MINUS(x)
+1(minus(+(x, 1)), 1) → MINUS(x)

The TRS R consists of the following rules:

minus(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(x)

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:

+1(x, minus(y)) → +1(minus(x), y)
+1(x, +(y, z)) → +1(x, y)
+1(x, minus(y)) → MINUS(+(minus(x), y))
+1(x, +(y, z)) → +1(+(x, y), z)
+1(x, minus(y)) → MINUS(x)
+1(minus(+(x, 1)), 1) → MINUS(x)

The TRS R consists of the following rules:

minus(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

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

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

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

The TRS R consists of the following rules:

minus(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(x)

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

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

POL(minus(x1)) = (11/4)x_1   
POL(0) = 2   
POL(1) = 13/4   
POL(+1(x1, x2)) = (1/4)x_2   
POL(+(x1, x2)) = 1/4 + (2)x_1 + (5/2)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
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof

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

+1(x, minus(y)) → +1(minus(x), y)

The TRS R consists of the following rules:

minus(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(x)

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

POL(minus(x1)) = 4 + (2)x_1   
POL(0) = 2   
POL(+1(x1, x2)) = (1/4)x_2   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



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

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

minus(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(x)

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.