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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
↳7 QDPOrderProof (⇔)
↳8 QDP
↳9 DependencyGraphProof (⇔)
↳10 QDP
↳11 QDPOrderProof (⇔)
↳12 QDP
↳13 DependencyGraphProof (⇔)
↳14 QDP
↳15 QDPOrderProof (⇔)
↳16 QDP
↳17 DependencyGraphProof (⇔)
↳18 QDP
↳19 UsableRulesProof (⇔)
↳20 QDP
↳21 QReductionProof (⇔)
↳22 QDP
↳23 UsableRulesReductionPairsProof (⇔)
↳24 QDP
↳25 QDPSizeChangeProof (⇔)
↳26 TRUE

(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.


U31(dout(DX), X, Y) → DIN(der(Y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented:

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

(8) 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)
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.

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) Obligation:

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

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

(11) 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))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(DIN(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(der(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(plus(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/01\
\00/
·x2

POL(U21(x1, x2, x3)) =
/0\
\0/
 +
/01\
\00/
·x1 +
/00\
\00/
·x2 +
/00\
\00/
·x3

POL(din(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(dout(x1)) =
/0\
\1/
 +
/11\
\01/
·x1

POL(times(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(U41(x1, x2)) =
/0\
\0/
 +
/00\
\10/
·x1 +
/00\
\00/
·x2

POL(u21(x1, x2, x3)) =
/0\
\0/
 +
/11\
\11/
·x1 +
/00\
\00/
·x2 +
/00\
\00/
·x3

POL(u41(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(u42(x1, x2, x3)) =
/0\
\0/
 +
/10\
\01/
·x1 +
/00\
\00/
·x2 +
/00\
\00/
·x3

POL(u32(x1, x2, x3, x4)) =
/0\
\0/
 +
/00\
\01/
·x1 +
/00\
\00/
·x2 +
/00\
\00/
·x3 +
/00\
\00/
·x4

POL(u31(x1, x2, x3)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2 +
/00\
\00/
·x3

POL(u22(x1, x2, x3, x4)) =
/0\
\1/
 +
/10\
\01/
·x1 +
/00\
\00/
·x2 +
/00\
\00/
·x3 +
/00\
\11/
·x4

The following usable rules [FROCOS05] were oriented:

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

(12) 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))) → 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.

(13) DependencyGraphProof (EQUIVALENT transformation)

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

(14) 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))) → 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.

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Polynomial interpretation [POLO]:

POL(DIN(x1)) = 0   
POL(U41(x1, x2)) = x1   
POL(der(x1)) = 0   
POL(din(x1)) = 0   
POL(dout(x1)) = 1   
POL(plus(x1, x2)) = 0   
POL(times(x1, x2)) = 0   
POL(u21(x1, x2, x3)) = x1   
POL(u22(x1, x2, x3, x4)) = 1   
POL(u31(x1, x2, x3)) = x1   
POL(u32(x1, x2, x3, x4)) = 1 + x1   
POL(u41(x1, x2)) = x1   
POL(u42(x1, x2, x3)) = 1   

The following usable rules [FROCOS05] were oriented:

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

(16) 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))) → 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.

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) Obligation:

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

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

(19) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(20) Obligation:

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

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

R is empty.
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.

(21) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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)

(22) Obligation:

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

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

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

(23) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

DIN(der(plus(X, Y))) → DIN(der(X))
DIN(der(times(X, Y))) → DIN(der(X))
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(DIN(x1)) = 2·x1   
POL(der(x1)) = 2·x1   
POL(plus(x1, x2)) = 2·x1 + x2   
POL(times(x1, x2)) = 2·x1 + x2   

(24) Obligation:

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

DIN(der(der(X))) → DIN(der(X))

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

(25) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • DIN(der(der(X))) → DIN(der(X))
    The graph contains the following edges 1 > 1

(26) TRUE