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:

din1(der1(plus2(X, Y))) -> u213(din1(der1(X)), X, Y)
u213(dout1(DX), X, Y) -> u224(din1(der1(Y)), X, Y, DX)
u224(dout1(DY), X, Y, DX) -> dout1(plus2(DX, DY))
din1(der1(times2(X, Y))) -> u313(din1(der1(X)), X, Y)
u313(dout1(DX), X, Y) -> u324(din1(der1(Y)), X, Y, DX)
u324(dout1(DY), X, Y, DX) -> dout1(plus2(times2(X, DY), times2(Y, DX)))
din1(der1(der1(X))) -> u412(din1(der1(X)), X)
u412(dout1(DX), X) -> u423(din1(der1(DX)), X, DX)
u423(dout1(DDX), X, DX) -> dout1(DDX)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

din1(der1(plus2(X, Y))) -> u213(din1(der1(X)), X, Y)
u213(dout1(DX), X, Y) -> u224(din1(der1(Y)), X, Y, DX)
u224(dout1(DY), X, Y, DX) -> dout1(plus2(DX, DY))
din1(der1(times2(X, Y))) -> u313(din1(der1(X)), X, Y)
u313(dout1(DX), X, Y) -> u324(din1(der1(Y)), X, Y, DX)
u324(dout1(DY), X, Y, DX) -> dout1(plus2(times2(X, DY), times2(Y, DX)))
din1(der1(der1(X))) -> u412(din1(der1(X)), X)
u412(dout1(DX), X) -> u423(din1(der1(DX)), X, DX)
u423(dout1(DDX), X, DX) -> dout1(DDX)

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:

U412(dout1(DX), X) -> U423(din1(der1(DX)), X, DX)
U213(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(plus2(X, Y))) -> U213(din1(der1(X)), X, Y)
DIN1(der1(der1(X))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
U313(dout1(DX), X, Y) -> U324(din1(der1(Y)), X, Y, DX)
DIN1(der1(plus2(X, Y))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> U313(din1(der1(X)), X, Y)
U213(dout1(DX), X, Y) -> U224(din1(der1(Y)), X, Y, DX)
U412(dout1(DX), X) -> DIN1(der1(DX))
U313(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(der1(X))) -> U412(din1(der1(X)), X)

The TRS R consists of the following rules:

din1(der1(plus2(X, Y))) -> u213(din1(der1(X)), X, Y)
u213(dout1(DX), X, Y) -> u224(din1(der1(Y)), X, Y, DX)
u224(dout1(DY), X, Y, DX) -> dout1(plus2(DX, DY))
din1(der1(times2(X, Y))) -> u313(din1(der1(X)), X, Y)
u313(dout1(DX), X, Y) -> u324(din1(der1(Y)), X, Y, DX)
u324(dout1(DY), X, Y, DX) -> dout1(plus2(times2(X, DY), times2(Y, DX)))
din1(der1(der1(X))) -> u412(din1(der1(X)), X)
u412(dout1(DX), X) -> u423(din1(der1(DX)), X, DX)
u423(dout1(DDX), X, DX) -> dout1(DDX)

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:

U412(dout1(DX), X) -> U423(din1(der1(DX)), X, DX)
U213(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(plus2(X, Y))) -> U213(din1(der1(X)), X, Y)
DIN1(der1(der1(X))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
U313(dout1(DX), X, Y) -> U324(din1(der1(Y)), X, Y, DX)
DIN1(der1(plus2(X, Y))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> U313(din1(der1(X)), X, Y)
U213(dout1(DX), X, Y) -> U224(din1(der1(Y)), X, Y, DX)
U412(dout1(DX), X) -> DIN1(der1(DX))
U313(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(der1(X))) -> U412(din1(der1(X)), X)

The TRS R consists of the following rules:

din1(der1(plus2(X, Y))) -> u213(din1(der1(X)), X, Y)
u213(dout1(DX), X, Y) -> u224(din1(der1(Y)), X, Y, DX)
u224(dout1(DY), X, Y, DX) -> dout1(plus2(DX, DY))
din1(der1(times2(X, Y))) -> u313(din1(der1(X)), X, Y)
u313(dout1(DX), X, Y) -> u324(din1(der1(Y)), X, Y, DX)
u324(dout1(DY), X, Y, DX) -> dout1(plus2(times2(X, DY), times2(Y, DX)))
din1(der1(der1(X))) -> u412(din1(der1(X)), X)
u412(dout1(DX), X) -> u423(din1(der1(DX)), X, DX)
u423(dout1(DDX), X, DX) -> dout1(DDX)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

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

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

DIN1(der1(plus2(X, Y))) -> U213(din1(der1(X)), X, Y)
U213(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(der1(X))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
DIN1(der1(plus2(X, Y))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> U313(din1(der1(X)), X, Y)
U412(dout1(DX), X) -> DIN1(der1(DX))
U313(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(der1(X))) -> U412(din1(der1(X)), X)

The TRS R consists of the following rules:

din1(der1(plus2(X, Y))) -> u213(din1(der1(X)), X, Y)
u213(dout1(DX), X, Y) -> u224(din1(der1(Y)), X, Y, DX)
u224(dout1(DY), X, Y, DX) -> dout1(plus2(DX, DY))
din1(der1(times2(X, Y))) -> u313(din1(der1(X)), X, Y)
u313(dout1(DX), X, Y) -> u324(din1(der1(Y)), X, Y, DX)
u324(dout1(DY), X, Y, DX) -> dout1(plus2(times2(X, DY), times2(Y, DX)))
din1(der1(der1(X))) -> u412(din1(der1(X)), X)
u412(dout1(DX), X) -> u423(din1(der1(DX)), X, DX)
u423(dout1(DDX), X, DX) -> dout1(DDX)

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.


U412(dout1(DX), X) -> DIN1(der1(DX))
The remaining pairs can at least be oriented weakly.

DIN1(der1(plus2(X, Y))) -> U213(din1(der1(X)), X, Y)
U213(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(der1(X))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
DIN1(der1(plus2(X, Y))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> U313(din1(der1(X)), X, Y)
U313(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(der1(X))) -> U412(din1(der1(X)), X)
Used ordering: Polynomial interpretation [21]:

POL(DIN1(x1)) = 0   
POL(U213(x1, x2, x3)) = 0   
POL(U313(x1, x2, x3)) = 0   
POL(U412(x1, x2)) = x1   
POL(der1(x1)) = 0   
POL(din1(x1)) = 0   
POL(dout1(x1)) = 1   
POL(plus2(x1, x2)) = 0   
POL(times2(x1, x2)) = 0   
POL(u213(x1, x2, x3)) = x1   
POL(u224(x1, x2, x3, x4)) = 1 + x1   
POL(u313(x1, x2, x3)) = x1   
POL(u324(x1, x2, x3, x4)) = x1   
POL(u412(x1, x2)) = x1   
POL(u423(x1, x2, x3)) = x1   

The following usable rules [14] were oriented:

din1(der1(der1(X))) -> u412(din1(der1(X)), X)
din1(der1(times2(X, Y))) -> u313(din1(der1(X)), X, Y)
u324(dout1(DY), X, Y, DX) -> dout1(plus2(times2(X, DY), times2(Y, DX)))
din1(der1(plus2(X, Y))) -> u213(din1(der1(X)), X, Y)
u423(dout1(DDX), X, DX) -> dout1(DDX)
u213(dout1(DX), X, Y) -> u224(din1(der1(Y)), X, Y, DX)
u313(dout1(DX), X, Y) -> u324(din1(der1(Y)), X, Y, DX)
u224(dout1(DY), X, Y, DX) -> dout1(plus2(DX, DY))
u412(dout1(DX), X) -> u423(din1(der1(DX)), X, DX)



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

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

U213(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(plus2(X, Y))) -> U213(din1(der1(X)), X, Y)
DIN1(der1(der1(X))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
DIN1(der1(plus2(X, Y))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> U313(din1(der1(X)), X, Y)
U313(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(der1(X))) -> U412(din1(der1(X)), X)

The TRS R consists of the following rules:

din1(der1(plus2(X, Y))) -> u213(din1(der1(X)), X, Y)
u213(dout1(DX), X, Y) -> u224(din1(der1(Y)), X, Y, DX)
u224(dout1(DY), X, Y, DX) -> dout1(plus2(DX, DY))
din1(der1(times2(X, Y))) -> u313(din1(der1(X)), X, Y)
u313(dout1(DX), X, Y) -> u324(din1(der1(Y)), X, Y, DX)
u324(dout1(DY), X, Y, DX) -> dout1(plus2(times2(X, DY), times2(Y, DX)))
din1(der1(der1(X))) -> u412(din1(der1(X)), X)
u412(dout1(DX), X) -> u423(din1(der1(DX)), X, DX)
u423(dout1(DDX), X, DX) -> dout1(DDX)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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

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

DIN1(der1(plus2(X, Y))) -> U213(din1(der1(X)), X, Y)
U213(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(der1(X))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
DIN1(der1(plus2(X, Y))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> U313(din1(der1(X)), X, Y)
U313(dout1(DX), X, Y) -> DIN1(der1(Y))

The TRS R consists of the following rules:

din1(der1(plus2(X, Y))) -> u213(din1(der1(X)), X, Y)
u213(dout1(DX), X, Y) -> u224(din1(der1(Y)), X, Y, DX)
u224(dout1(DY), X, Y, DX) -> dout1(plus2(DX, DY))
din1(der1(times2(X, Y))) -> u313(din1(der1(X)), X, Y)
u313(dout1(DX), X, Y) -> u324(din1(der1(Y)), X, Y, DX)
u324(dout1(DY), X, Y, DX) -> dout1(plus2(times2(X, DY), times2(Y, DX)))
din1(der1(der1(X))) -> u412(din1(der1(X)), X)
u412(dout1(DX), X) -> u423(din1(der1(DX)), X, DX)
u423(dout1(DDX), X, DX) -> dout1(DDX)

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.


U213(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
DIN1(der1(plus2(X, Y))) -> DIN1(der1(X))
U313(dout1(DX), X, Y) -> DIN1(der1(Y))
The remaining pairs can at least be oriented weakly.

DIN1(der1(plus2(X, Y))) -> U213(din1(der1(X)), X, Y)
DIN1(der1(der1(X))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> U313(din1(der1(X)), X, Y)
Used ordering: Polynomial interpretation [21]:

POL(DIN1(x1)) = x1   
POL(U213(x1, x2, x3)) = 1 + x3   
POL(U313(x1, x2, x3)) = 1 + x3   
POL(der1(x1)) = x1   
POL(din1(x1)) = 0   
POL(dout1(x1)) = 0   
POL(plus2(x1, x2)) = 1 + x1 + x2   
POL(times2(x1, x2)) = 1 + x1 + x2   
POL(u213(x1, x2, x3)) = 0   
POL(u224(x1, x2, x3, x4)) = 0   
POL(u313(x1, x2, x3)) = 0   
POL(u324(x1, x2, x3, x4)) = 0   
POL(u412(x1, x2)) = 0   
POL(u423(x1, x2, x3)) = 0   

The following usable rules [14] were oriented: none



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

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

DIN1(der1(plus2(X, Y))) -> U213(din1(der1(X)), X, Y)
DIN1(der1(der1(X))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> U313(din1(der1(X)), X, Y)

The TRS R consists of the following rules:

din1(der1(plus2(X, Y))) -> u213(din1(der1(X)), X, Y)
u213(dout1(DX), X, Y) -> u224(din1(der1(Y)), X, Y, DX)
u224(dout1(DY), X, Y, DX) -> dout1(plus2(DX, DY))
din1(der1(times2(X, Y))) -> u313(din1(der1(X)), X, Y)
u313(dout1(DX), X, Y) -> u324(din1(der1(Y)), X, Y, DX)
u324(dout1(DY), X, Y, DX) -> dout1(plus2(times2(X, DY), times2(Y, DX)))
din1(der1(der1(X))) -> u412(din1(der1(X)), X)
u412(dout1(DX), X) -> u423(din1(der1(DX)), X, DX)
u423(dout1(DDX), X, DX) -> dout1(DDX)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

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

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

DIN1(der1(der1(X))) -> DIN1(der1(X))

The TRS R consists of the following rules:

din1(der1(plus2(X, Y))) -> u213(din1(der1(X)), X, Y)
u213(dout1(DX), X, Y) -> u224(din1(der1(Y)), X, Y, DX)
u224(dout1(DY), X, Y, DX) -> dout1(plus2(DX, DY))
din1(der1(times2(X, Y))) -> u313(din1(der1(X)), X, Y)
u313(dout1(DX), X, Y) -> u324(din1(der1(Y)), X, Y, DX)
u324(dout1(DY), X, Y, DX) -> dout1(plus2(times2(X, DY), times2(Y, DX)))
din1(der1(der1(X))) -> u412(din1(der1(X)), X)
u412(dout1(DX), X) -> u423(din1(der1(DX)), X, DX)
u423(dout1(DDX), X, DX) -> dout1(DDX)

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.


DIN1(der1(der1(X))) -> DIN1(der1(X))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(DIN1(x1)) = x1   
POL(der1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

din1(der1(plus2(X, Y))) -> u213(din1(der1(X)), X, Y)
u213(dout1(DX), X, Y) -> u224(din1(der1(Y)), X, Y, DX)
u224(dout1(DY), X, Y, DX) -> dout1(plus2(DX, DY))
din1(der1(times2(X, Y))) -> u313(din1(der1(X)), X, Y)
u313(dout1(DX), X, Y) -> u324(din1(der1(Y)), X, Y, DX)
u324(dout1(DY), X, Y, DX) -> dout1(plus2(times2(X, DY), times2(Y, DX)))
din1(der1(der1(X))) -> u412(din1(der1(X)), X)
u412(dout1(DX), X) -> u423(din1(der1(DX)), X, DX)
u423(dout1(DDX), X, DX) -> dout1(DDX)

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.