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, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(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:

F(h(x, h(y, z))) → +1(x, y)
+1(x, +(y, z)) → +1(x, y)
F(g(h(x, y))) → F(h(s(x), y))
F(g(f(x))) → F(h(s(0), x))
+1(x, +(y, z)) → +1(+(x, y), z)
+1(x, s(y)) → +1(x, y)
F(h(x, h(y, z))) → F(h(+(x, y), z))
+1(s(x), y) → +1(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, 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(h(x, h(y, z))) → +1(x, y)
+1(x, +(y, z)) → +1(x, y)
F(g(h(x, y))) → F(h(s(x), y))
F(g(f(x))) → F(h(s(0), x))
+1(x, +(y, z)) → +1(+(x, y), z)
+1(x, s(y)) → +1(x, y)
F(h(x, h(y, z))) → F(h(+(x, y), z))
+1(s(x), y) → +1(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, 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 2 SCCs with 3 less nodes.

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

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

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

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(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(x, y)
+1(x, s(y)) → +1(x, y)
+1(s(x), y) → +1(x, y)
The remaining pairs can at least be oriented weakly.

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

POL(s(x1)) = 4 + x_1   
POL(0) = 0   
POL(+1(x1, x2)) = (4)x_1 + (4)x_2   
POL(+(x1, x2)) = 3 + x_1 + x_2   
The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)



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

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

+1(x, +(y, z)) → +1(+(x, y), z)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(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(+(x, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 1 + (4)x_1   
POL(0) = 0   
POL(+1(x1, x2)) = (3)x_2   
POL(+(x1, x2)) = 3 + (3)x_2   
The value of delta used in the strict ordering is 9.
The following usable rules [17] were oriented: none



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

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(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.

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

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

F(h(x, h(y, z))) → F(h(+(x, y), z))

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(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.


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

POL(h(x1, x2)) = 3 + (3)x_2   
POL(s(x1)) = 2 + (2)x_1   
POL(0) = 0   
POL(F(x1)) = (3)x_1   
POL(+(x1, x2)) = 0   
The value of delta used in the strict ordering is 27.
The following usable rules [17] were oriented: none



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

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(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.