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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

+1(a, b) → +1(b, a)
F(+(x, y), z) → F(y, z)
+1(+(x, y), z) → +1(y, z)
+1(a, +(b, z)) → +1(a, z)
+1(a, +(b, z)) → +1(b, +(a, z))
F(+(x, y), z) → F(x, z)
F(+(x, y), z) → +1(f(x, z), f(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 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(a, b) → +1(b, a)
F(+(x, y), z) → F(y, z)
+1(+(x, y), z) → +1(y, z)
+1(a, +(b, z)) → +1(a, z)
+1(a, +(b, z)) → +1(b, +(a, z))
F(+(x, y), z) → F(x, z)
+1(+(x, y), z) → +1(x, +(y, z))
F(+(x, y), z) → +1(f(x, z), f(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.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

A(B(z)) → A(z)

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


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: Combined order from the following AFS and order.
A(x1)  =  A(x1)
B(x1)  =  B(x1)

Recursive path order with status [2].
Precedence:
B1 > A1

Status:
A1: multiset
B1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
+1(x1, x2)  =  x1
+(x1, x2)  =  +(x1, x2)
a  =  a
b  =  b

Recursive path order with status [2].
Precedence:
a > +2
b > +2

Status:
b: multiset
a: multiset
+2: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
F(x1, x2)  =  F(x1)
+(x1, x2)  =  +(x1, x2)

Recursive path order with status [2].
Precedence:
+2 > F1

Status:
+2: multiset
F1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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.