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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)


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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)

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
  ↳ Non-Overlap Check
    ↳ 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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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

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)
U412(dout1(DX), X) -> DIN1(der1(DX))
U313(dout1(DX), X, Y) -> DIN1(der1(Y))
DIN1(der1(der1(X))) -> U412(din1(der1(X)), X)
Used ordering: Combined order from the following AFS and order.
DIN1(x1)  =  DIN
der1(x1)  =  der
plus2(x1, x2)  =  x2
U213(x1, x2, x3)  =  x1
din1(x1)  =  din
dout1(x1)  =  dout1(x1)
times2(x1, x2)  =  times
U313(x1, x2, x3)  =  U31
U412(x1, x2)  =  U41
u423(x1, x2, x3)  =  u421(x1)
u313(x1, x2, x3)  =  x1
u412(x1, x2)  =  x1
u324(x1, x2, x3, x4)  =  u321(x1)
u213(x1, x2, x3)  =  x1
u224(x1, x2, x3, x4)  =  x1

Lexicographic Path Order [19].
Precedence:
der > [DIN, din, dout1, U31, U41] > [times, u421, u321]


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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))) -> 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))
U412(dout1(DX), X) -> DIN1(der1(DX))
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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)

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
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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))
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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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

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))
DIN1(der1(der1(X))) -> U412(din1(der1(X)), X)
Used ordering: Combined order from the following AFS and order.
DIN1(x1)  =  DIN
der1(x1)  =  x1
times2(x1, x2)  =  x1
plus2(x1, x2)  =  plus
U313(x1, x2, x3)  =  x1
din1(x1)  =  din
U412(x1, x2)  =  U41
dout1(x1)  =  dout
u423(x1, x2, x3)  =  x1
u313(x1, x2, x3)  =  u31
u412(x1, x2)  =  u41
u324(x1, x2, x3, x4)  =  x1
u213(x1, x2, x3)  =  u21
u224(x1, x2, x3, x4)  =  x1

Lexicographic Path Order [19].
Precedence:
dout > [DIN, plus, din, U41, u31, u41, u21]


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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(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))
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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)

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
  ↳ Non-Overlap Check
    ↳ 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))
DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
DIN1(der1(plus2(X, Y))) -> DIN1(der1(X))
U412(dout1(DX), X) -> DIN1(der1(DX))
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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


U412(dout1(DX), X) -> DIN1(der1(DX))
DIN1(der1(der1(X))) -> U412(din1(der1(X)), X)
The remaining pairs can at least by weakly be oriented.

DIN1(der1(der1(X))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
DIN1(der1(plus2(X, Y))) -> DIN1(der1(X))
Used ordering: Combined order from the following AFS and order.
DIN1(x1)  =  DIN
der1(x1)  =  x1
times2(x1, x2)  =  x2
plus2(x1, x2)  =  plus2(x1, x2)
U412(x1, x2)  =  U411(x1)
dout1(x1)  =  dout
din1(x1)  =  din
u423(x1, x2, x3)  =  u42
u313(x1, x2, x3)  =  x1
u213(x1, x2, x3)  =  u21
u224(x1, x2, x3, x4)  =  x1
u324(x1, x2, x3, x4)  =  u32
u412(x1, x2)  =  x1

Lexicographic Path Order [19].
Precedence:
[DIN, dout, u42, u32] > [plus2, U411, din, u21]


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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:

DIN1(der1(der1(X))) -> DIN1(der1(X))
DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
DIN1(der1(plus2(X, Y))) -> 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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


DIN1(der1(der1(X))) -> DIN1(der1(X))
The remaining pairs can at least by weakly be oriented.

DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
DIN1(der1(plus2(X, Y))) -> DIN1(der1(X))
Used ordering: Combined order from the following AFS and order.
DIN1(x1)  =  x1
der1(x1)  =  der1(x1)
times2(x1, x2)  =  x1
plus2(x1, x2)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ QDPOrderProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ QDPOrderProof
                                ↳ QDP
                                  ↳ QDPOrderProof
QDP
                                      ↳ QDPOrderProof

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

DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
DIN1(der1(plus2(X, Y))) -> 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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


DIN1(der1(plus2(X, Y))) -> DIN1(der1(X))
The remaining pairs can at least by weakly be oriented.

DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
Used ordering: Combined order from the following AFS and order.
DIN1(x1)  =  DIN1(x1)
der1(x1)  =  x1
times2(x1, x2)  =  x1
plus2(x1, x2)  =  plus1(x1)

Lexicographic Path Order [19].
Precedence:
plus1 > DIN1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ QDPOrderProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ QDPOrderProof
                                ↳ QDP
                                  ↳ QDPOrderProof
                                    ↳ QDP
                                      ↳ QDPOrderProof
QDP
                                          ↳ QDPOrderProof

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

DIN1(der1(times2(X, Y))) -> 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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


DIN1(der1(times2(X, Y))) -> DIN1(der1(X))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DIN1(x1)  =  DIN1(x1)
der1(x1)  =  der1(x1)
times2(x1, x2)  =  times1(x1)

Lexicographic Path Order [19].
Precedence:
[DIN1, der1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ QDPOrderProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ QDPOrderProof
                                ↳ QDP
                                  ↳ QDPOrderProof
                                    ↳ QDP
                                      ↳ QDPOrderProof
                                        ↳ 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)

The set Q consists of the following terms:

din1(der1(plus2(x0, x1)))
u213(dout1(x0), x1, x2)
u224(dout1(x0), x1, x2, x3)
din1(der1(times2(x0, x1)))
u313(dout1(x0), x1, x2)
u324(dout1(x0), x1, x2, x3)
din1(der1(der1(x0)))
u412(dout1(x0), x1)
u423(dout1(x0), x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.