Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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)

The set Q consists of the following terms:

din(der(plus(x0, x1)))
u21(dout(x0), x1, x2)
u22(dout(x0), x1, x2, x3)
din(der(times(x0, x1)))
u31(dout(x0), x1, x2)
u32(dout(x0), x1, x2, x3)
din(der(der(x0)))
u41(dout(x0), x1)
u42(dout(x0), x1, x2)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

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)

The set Q consists of the following terms:

din(der(plus(x0, x1)))
u21(dout(x0), x1, x2)
u22(dout(x0), x1, x2, x3)
din(der(times(x0, x1)))
u31(dout(x0), x1, x2)
u32(dout(x0), x1, x2, x3)
din(der(der(x0)))
u41(dout(x0), x1)
u42(dout(x0), x1, x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

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

The set Q consists of the following terms:

din(der(plus(x0, x1)))
u21(dout(x0), x1, x2)
u22(dout(x0), x1, x2, x3)
din(der(times(x0, x1)))
u31(dout(x0), x1, x2)
u32(dout(x0), x1, x2, x3)
din(der(der(x0)))
u41(dout(x0), x1)
u42(dout(x0), x1, x2)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U21(x1, x2, x3)  =  U21(x1)
dout(x1)  =  dout
DIN(x1)  =  DIN(x1)
der(x1)  =  der
plus(x1, x2)  =  x2
din(x1)  =  x1
times(x1, x2)  =  x2
U31(x1, x2, x3)  =  U31(x1)
U41(x1, x2)  =  U41(x1)
u21(x1, x2, x3)  =  u21
u31(x1, x2, x3)  =  x1
u41(x1, x2)  =  x1
u42(x1, x2, x3)  =  x1
u22(x1, x2, x3, x4)  =  x1
u32(x1, x2, x3, x4)  =  u32(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

dout > [U21₁, DIN₁, der, U31₁, U41₁, u21, u32₁]

Status:

U21₁: [1]
dout: []
DIN₁: [1]
der: []
U31₁: [1]
U41₁: [1]
u21: []
u32₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(8) Obligation:

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

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

The set Q consists of the following terms:

din(der(plus(x0, x1)))
u21(dout(x0), x1, x2)
u22(dout(x0), x1, x2, x3)
din(der(times(x0, x1)))
u31(dout(x0), x1, x2)
u32(dout(x0), x1, x2, x3)
din(der(der(x0)))
u41(dout(x0), x1)
u42(dout(x0), x1, x2)

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(10) Obligation:

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

DIN(der(times(X, Y))) → DIN(der(X))
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)

The set Q consists of the following terms:

din(der(plus(x0, x1)))
u21(dout(x0), x1, x2)
u22(dout(x0), x1, x2, x3)
din(der(times(x0, x1)))
u31(dout(x0), x1, x2)
u32(dout(x0), x1, x2, x3)
din(der(der(x0)))
u41(dout(x0), x1)
u42(dout(x0), x1, x2)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIN(x1)  =  x1
der(x1)  =  der(x1)
times(x1, x2)  =  times(x1)
plus(x1, x2)  =  plus(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

times₁ > [der₁, plus₁]

Status:

der₁: [1]
times₁: [1]
plus₁: [1]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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)

The set Q consists of the following terms:

din(der(plus(x0, x1)))
u21(dout(x0), x1, x2)
u22(dout(x0), x1, x2, x3)
din(der(times(x0, x1)))
u31(dout(x0), x1, x2)
u32(dout(x0), x1, x2, x3)
din(der(der(x0)))
u41(dout(x0), x1)
u42(dout(x0), x1, x2)

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

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE