Termination w.r.t. Q proof of TRS/Rubiowst99.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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.