Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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 [1,13] 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(f(x), +(f(y), z)) → +1(x, y)
+1(+(x, y), z) → +1(y, z)
+1(f(x), +(f(y), z)) → +1(f(+(x, 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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

+1(f(x), f(y)) → +1(x, y)
+1(f(x), +(f(y), z)) → +1(x, y)
+1(+(x, y), z) → +1(y, z)
+1(f(x), +(f(y), z)) → +1(f(+(x, 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 deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP

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