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:

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))

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, f(a, y)) → F(a, x)
F(x, f(a, y)) → F(a, f(f(a, h(f(a, x))), y))
F(x, f(a, y)) → F(a, h(f(a, x)))
F(x, f(a, y)) → F(f(a, h(f(a, x))), y)

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))

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, f(a, y)) → F(a, x)
F(x, f(a, y)) → F(a, f(f(a, h(f(a, x))), y))
F(x, f(a, y)) → F(a, h(f(a, x)))
F(x, f(a, y)) → F(f(a, h(f(a, x))), y)

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))

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 1 less node.

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

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

F(x, f(a, y)) → F(a, x)
F(x, f(a, y)) → F(a, f(f(a, h(f(a, x))), y))
F(x, f(a, y)) → F(f(a, h(f(a, x))), y)

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))

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, f(a, y)) → F(a, x)
The remaining pairs can at least be oriented weakly.

F(x, f(a, y)) → F(a, f(f(a, h(f(a, x))), y))
F(x, f(a, y)) → F(f(a, h(f(a, x))), y)
Used ordering: Polynomial interpretation [25]:

POL(F(x1, x2)) = x1 + x2   
POL(a) = 0   
POL(f(x1, x2)) = 1 + x2   
POL(h(x1)) = 0   

The following usable rules [17] were oriented:

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))



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

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

F(x, f(a, y)) → F(a, f(f(a, h(f(a, x))), y))
F(x, f(a, y)) → F(f(a, h(f(a, x))), y)

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))

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, f(a, y)) → F(f(a, h(f(a, x))), y)
The remaining pairs can at least be oriented weakly.

F(x, f(a, y)) → F(a, f(f(a, h(f(a, x))), y))
Used ordering: Polynomial interpretation with max and min functions [25]:

POL(F(x1, x2)) = x2   
POL(a) = 0   
POL(f(x1, x2)) = 1 + x2   
POL(h(x1)) = 0   

The following usable rules [17] were oriented:

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))



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

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

F(x, f(a, y)) → F(a, f(f(a, h(f(a, x))), y))

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(x, f(a, y)) → F(a, f(f(a, h(f(a, x))), y)) we obtained the following new rules:

F(a, f(a, x1)) → F(a, f(f(a, h(f(a, a))), x1))



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

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

F(a, f(a, x1)) → F(a, f(f(a, h(f(a, a))), x1))

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))

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(a, f(a, x1)) → F(a, f(f(a, h(f(a, a))), x1))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( a ) =
/1\
\1/

M( f(x1, x2) ) =
/0\
\1/
+
/00\
\10/
·x1+
/00\
\01/
·x2

M( h(x1) ) =
/0\
\0/
+
/00\
\01/
·x1

Tuple symbols:
M( F(x1, x2) ) = 0+
[1,1]
·x1+
[0,1]
·x2


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))



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

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

f(x, f(a, y)) → f(a, f(f(a, h(f(a, x))), y))

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.