Termination w.r.t. Q proof of /tmp/tmpccFtF1/wst99.trs

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:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

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:

DIN(der(times(X, Y))) → U31(din(der(X)), X, Y)
DIN(der(times(X, Y))) → DIN(der(X))
U41(dout(DX), X) → U42(din(der(DX)), X, DX)
DIN(der(der(X))) → U41(din(der(X)), X)
U31(dout(DX), X, Y) → U32(din(der(Y)), X, Y, DX)
U21(dout(DX), X, Y) → U22(din(der(Y)), X, Y, DX)
U41(dout(DX), X) → DIN(der(DX))
DIN(der(plus(X, Y))) → DIN(der(X))
U21(dout(DX), X, Y) → DIN(der(Y))
DIN(der(plus(X, Y))) → U21(din(der(X)), X, Y)
DIN(der(der(X))) → DIN(der(X))
U31(dout(DX), X, Y) → DIN(der(Y))

The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(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:

DIN(der(times(X, Y))) → U31(din(der(X)), X, Y)
DIN(der(times(X, Y))) → DIN(der(X))
U41(dout(DX), X) → U42(din(der(DX)), X, DX)
DIN(der(der(X))) → U41(din(der(X)), X)
U31(dout(DX), X, Y) → U32(din(der(Y)), X, Y, DX)
U21(dout(DX), X, Y) → U22(din(der(Y)), X, Y, DX)
U41(dout(DX), X) → DIN(der(DX))
DIN(der(plus(X, Y))) → DIN(der(X))
U21(dout(DX), X, Y) → DIN(der(Y))
DIN(der(plus(X, Y))) → U21(din(der(X)), X, Y)
DIN(der(der(X))) → DIN(der(X))
U31(dout(DX), X, Y) → DIN(der(Y))

The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

DIN(der(times(X, Y))) → U31(din(der(X)), X, Y)
DIN(der(times(X, Y))) → DIN(der(X))
DIN(der(der(X))) → U41(din(der(X)), X)
U41(dout(DX), X) → DIN(der(DX))
DIN(der(plus(X, Y))) → DIN(der(X))
DIN(der(plus(X, Y))) → U21(din(der(X)), X, Y)
U21(dout(DX), X, Y) → DIN(der(Y))
DIN(der(der(X))) → DIN(der(X))
U31(dout(DX), X, Y) → DIN(der(Y))

The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

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.

U41(dout(DX), X) → DIN(der(DX))

The remaining pairs can at least be oriented weakly.

DIN(der(times(X, Y))) → U31(din(der(X)), X, Y)
DIN(der(times(X, Y))) → DIN(der(X))
DIN(der(der(X))) → U41(din(der(X)), X)
DIN(der(plus(X, Y))) → DIN(der(X))
DIN(der(plus(X, Y))) → U21(din(der(X)), X, Y)
U21(dout(DX), X, Y) → DIN(der(Y))
DIN(der(der(X))) → DIN(der(X))
U31(dout(DX), X, Y) → DIN(der(Y))

Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = 0
POL(u32(x₁, x₂, x₃, x₄)) = (2)x_1
POL(u42(x₁, x₂, x₃)) = x_1
POL(u31(x₁, x₂, x₃)) = 0
POL(DIN(x₁)) = 0
POL(der(x₁)) = 0
POL(U31(x₁, x₂, x₃)) = 0
POL(din(x₁)) = 0
POL(U41(x₁, x₂)) = (2)x_1
POL(times(x₁, x₂)) = 0
POL(u22(x₁, x₂, x₃, x₄)) = (4)x_1
POL(u41(x₁, x₂)) = 0
POL(dout(x₁)) = 1/2
POL(U21(x₁, x₂, x₃)) = 0
POL(u21(x₁, x₂, x₃)) = 0

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
din(der(der(X))) → u41(din(der(X)), X)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
u42(dout(DDX), X, DX) → dout(DDX)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

DIN(der(times(X, Y))) → U31(din(der(X)), X, Y)
DIN(der(times(X, Y))) → DIN(der(X))
DIN(der(der(X))) → U41(din(der(X)), X)
DIN(der(plus(X, Y))) → DIN(der(X))
U21(dout(DX), X, Y) → DIN(der(Y))
DIN(der(plus(X, Y))) → U21(din(der(X)), X, Y)
DIN(der(der(X))) → DIN(der(X))
U31(dout(DX), X, Y) → DIN(der(Y))

The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

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

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

DIN(der(times(X, Y))) → U31(din(der(X)), X, Y)
DIN(der(times(X, Y))) → DIN(der(X))
DIN(der(plus(X, Y))) → DIN(der(X))
U21(dout(DX), X, Y) → DIN(der(Y))
DIN(der(plus(X, Y))) → U21(din(der(X)), X, Y)
DIN(der(der(X))) → DIN(der(X))
U31(dout(DX), X, Y) → DIN(der(Y))

The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

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.

DIN(der(times(X, Y))) → U31(din(der(X)), X, Y)
DIN(der(times(X, Y))) → DIN(der(X))

The remaining pairs can at least be oriented weakly.

DIN(der(plus(X, Y))) → DIN(der(X))
U21(dout(DX), X, Y) → DIN(der(Y))
DIN(der(plus(X, Y))) → U21(din(der(X)), X, Y)
DIN(der(der(X))) → DIN(der(X))
U31(dout(DX), X, Y) → DIN(der(Y))

Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = (4)x_1 + (2)x_2
POL(u32(x₁, x₂, x₃, x₄)) = 0
POL(u42(x₁, x₂, x₃)) = 0
POL(u31(x₁, x₂, x₃)) = 0
POL(DIN(x₁)) = (1/2)x_1
POL(der(x₁)) = (2)x_1
POL(U31(x₁, x₂, x₃)) = (4)x_3
POL(din(x₁)) = 0
POL(times(x₁, x₂)) = 4 + (4)x_1 + (4)x_2
POL(u22(x₁, x₂, x₃, x₄)) = 0
POL(u41(x₁, x₂)) = 0
POL(dout(x₁)) = 0
POL(U21(x₁, x₂, x₃)) = (2)x_3
POL(u21(x₁, x₂, x₃)) = 0

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

DIN(der(plus(X, Y))) → DIN(der(X))
DIN(der(plus(X, Y))) → U21(din(der(X)), X, Y)
U21(dout(DX), X, Y) → DIN(der(Y))
DIN(der(der(X))) → DIN(der(X))
U31(dout(DX), X, Y) → DIN(der(Y))

The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

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

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

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

DIN(der(plus(X, Y))) → DIN(der(X))
U21(dout(DX), X, Y) → DIN(der(Y))
DIN(der(plus(X, Y))) → U21(din(der(X)), X, Y)
DIN(der(der(X))) → DIN(der(X))

The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

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.

U21(dout(DX), X, Y) → DIN(der(Y))

The remaining pairs can at least be oriented weakly.

DIN(der(plus(X, Y))) → DIN(der(X))
DIN(der(plus(X, Y))) → U21(din(der(X)), X, Y)
DIN(der(der(X))) → DIN(der(X))

Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = 0
POL(u32(x₁, x₂, x₃, x₄)) = (4)x_1 + x_4
POL(u42(x₁, x₂, x₃)) = (4)x_1
POL(u31(x₁, x₂, x₃)) = (4)x_1
POL(DIN(x₁)) = 0
POL(der(x₁)) = 0
POL(din(x₁)) = 0
POL(times(x₁, x₂)) = 0
POL(u22(x₁, x₂, x₃, x₄)) = x_1
POL(u41(x₁, x₂)) = 0
POL(dout(x₁)) = 1/4 + (1/4)x_1
POL(U21(x₁, x₂, x₃)) = (1/4)x_1
POL(u21(x₁, x₂, x₃)) = 0

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
din(der(der(X))) → u41(din(der(X)), X)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
u42(dout(DDX), X, DX) → dout(DDX)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

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

DIN(der(plus(X, Y))) → DIN(der(X))
DIN(der(plus(X, Y))) → U21(din(der(X)), X, Y)
DIN(der(der(X))) → DIN(der(X))

The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

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

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

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

DIN(der(plus(X, Y))) → DIN(der(X))
DIN(der(der(X))) → DIN(der(X))

The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

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.

DIN(der(der(X))) → DIN(der(X))

The remaining pairs can at least be oriented weakly.

DIN(der(plus(X, Y))) → DIN(der(X))

Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = (4)x_1 + (1/4)x_2
POL(DIN(x₁)) = (1/4)x_1
POL(der(x₁)) = 1/4 + x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ QDPOrderProof

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

DIN(der(plus(X, Y))) → DIN(der(X))

The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

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.

DIN(der(plus(X, Y))) → DIN(der(X))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = 1 + x_1 + (4)x_2
POL(DIN(x₁)) = (1/4)x_1
POL(der(x₁)) = (1/4)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ PisEmptyProof

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

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(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.