Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

__(__(X, Y), Z) → __(X, __(Y, Z))
__(X, nil) → X
__(nil, X) → X
U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U61(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU61(isQid)
isNePalU71(isQid)
isPalU81(isNePal)
isPaltt
isQidtt

Q is empty.


QTRS
  ↳ RRRPoloQTRSProof

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

__(__(X, Y), Z) → __(X, __(Y, Z))
__(X, nil) → X
__(nil, X) → X
U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U61(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU61(isQid)
isNePalU71(isQid)
isPalU81(isNePal)
isPaltt
isQidtt

Q is empty.

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

__(__(X, Y), Z) → __(X, __(Y, Z))
__(X, nil) → X
__(nil, X) → X
U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U61(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU61(isQid)
isNePalU71(isQid)
isPalU81(isNePal)
isPaltt
isQidtt

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

__(__(X, Y), Z) → __(X, __(Y, Z))
__(X, nil) → X
__(nil, X) → X
Used ordering:
Polynomial interpretation [25]:

POL(U11(x1)) = 2·x1   
POL(U21(x1)) = 2·x1   
POL(U22(x1)) = 2·x1   
POL(U31(x1)) = 2·x1   
POL(U41(x1)) = 2·x1   
POL(U42(x1)) = x1   
POL(U51(x1)) = x1   
POL(U52(x1)) = x1   
POL(U61(x1)) = x1   
POL(U71(x1)) = 2·x1   
POL(U72(x1)) = 2·x1   
POL(U81(x1)) = 2·x1   
POL(__(x1, x2)) = 2 + 2·x1 + x2   
POL(isList) = 0   
POL(isNeList) = 0   
POL(isNePal) = 0   
POL(isPal) = 0   
POL(isQid) = 0   
POL(nil) = 1   
POL(tt) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
QTRS
      ↳ RRRPoloQTRSProof

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

U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U61(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU61(isQid)
isNePalU71(isQid)
isPalU81(isNePal)
isPaltt
isQidtt

Q is empty.

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

U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U61(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU61(isQid)
isNePalU71(isQid)
isPalU81(isNePal)
isPaltt
isQidtt

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

isNePalU61(isQid)
isPaltt
Used ordering:
Polynomial interpretation [25]:

POL(U11(x1)) = 2·x1   
POL(U21(x1)) = 2·x1   
POL(U22(x1)) = x1   
POL(U31(x1)) = x1   
POL(U41(x1)) = 2·x1   
POL(U42(x1)) = 2·x1   
POL(U51(x1)) = 2·x1   
POL(U52(x1)) = 2·x1   
POL(U61(x1)) = 2·x1   
POL(U71(x1)) = 1 + 2·x1   
POL(U72(x1)) = x1   
POL(U81(x1)) = x1   
POL(isList) = 0   
POL(isNeList) = 0   
POL(isNePal) = 1   
POL(isPal) = 1   
POL(isQid) = 0   
POL(tt) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
QTRS
          ↳ RRRPoloQTRSProof

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

U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U61(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU71(isQid)
isPalU81(isNePal)
isQidtt

Q is empty.

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

U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U61(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU71(isQid)
isPalU81(isNePal)
isQidtt

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

U61(tt) → tt
Used ordering:
Polynomial interpretation [25]:

POL(U11(x1)) = 2·x1   
POL(U21(x1)) = 2·x1   
POL(U22(x1)) = x1   
POL(U31(x1)) = x1   
POL(U41(x1)) = x1   
POL(U42(x1)) = 2·x1   
POL(U51(x1)) = 2·x1   
POL(U52(x1)) = 2·x1   
POL(U61(x1)) = 1 + x1   
POL(U71(x1)) = 2·x1   
POL(U72(x1)) = x1   
POL(U81(x1)) = 2·x1   
POL(isList) = 0   
POL(isNeList) = 0   
POL(isNePal) = 0   
POL(isPal) = 0   
POL(isQid) = 0   
POL(tt) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
QTRS
              ↳ DependencyPairsProof

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

U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU71(isQid)
isPalU81(isNePal)
isQidtt

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:

ISPALISNEPAL
U711(tt) → U721(isPal)
ISNELISTISNELIST
ISNELISTISLIST
ISLISTISLIST
U711(tt) → ISPAL
ISLISTISNELIST
U211(tt) → U221(isList)
U411(tt) → U421(isNeList)
U511(tt) → U521(isList)
ISNELISTU411(isList)
ISPALU811(isNePal)
ISNEPALU711(isQid)
U211(tt) → ISLIST
ISNELISTU511(isNeList)
U511(tt) → ISLIST
U411(tt) → ISNELIST
ISLISTU111(isNeList)
ISNELISTU311(isQid)
ISNELISTISQID
ISNEPALISQID
ISLISTU211(isList)

The TRS R consists of the following rules:

U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU71(isQid)
isPalU81(isNePal)
isQidtt

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
QDP
                  ↳ DependencyGraphProof

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

ISPALISNEPAL
U711(tt) → U721(isPal)
ISNELISTISNELIST
ISNELISTISLIST
ISLISTISLIST
U711(tt) → ISPAL
ISLISTISNELIST
U211(tt) → U221(isList)
U411(tt) → U421(isNeList)
U511(tt) → U521(isList)
ISNELISTU411(isList)
ISPALU811(isNePal)
ISNEPALU711(isQid)
U211(tt) → ISLIST
ISNELISTU511(isNeList)
U511(tt) → ISLIST
U411(tt) → ISNELIST
ISLISTU111(isNeList)
ISNELISTU311(isQid)
ISNELISTISQID
ISNEPALISQID
ISLISTU211(isList)

The TRS R consists of the following rules:

U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU71(isQid)
isPalU81(isNePal)
isQidtt

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 9 less nodes.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
QDP
                        ↳ UsableRulesProof
                      ↳ QDP

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

ISPALISNEPAL
U711(tt) → ISPAL
ISNEPALU711(isQid)

The TRS R consists of the following rules:

U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU71(isQid)
isPalU81(isNePal)
isQidtt

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
QDP
                            ↳ MNOCProof
                      ↳ QDP

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

ISPALISNEPAL
U711(tt) → ISPAL
ISNEPALU711(isQid)

The TRS R consists of the following rules:

isQidtt

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ MNOCProof
QDP
                                ↳ Rewriting
                      ↳ QDP

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

ISPALISNEPAL
U711(tt) → ISPAL
ISNEPALU711(isQid)

The TRS R consists of the following rules:

isQidtt

The set Q consists of the following terms:

isQid

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule ISNEPALU711(isQid) at position [0] we obtained the following new rules:

ISNEPALU711(tt)



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ MNOCProof
                              ↳ QDP
                                ↳ Rewriting
QDP
                                    ↳ UsableRulesProof
                      ↳ QDP

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

ISPALISNEPAL
U711(tt) → ISPAL
ISNEPALU711(tt)

The TRS R consists of the following rules:

isQidtt

The set Q consists of the following terms:

isQid

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ MNOCProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ UsableRulesProof
QDP
                                        ↳ QReductionProof
                      ↳ QDP

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

ISPALISNEPAL
U711(tt) → ISPAL
ISNEPALU711(tt)

R is empty.
The set Q consists of the following terms:

isQid

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

isQid



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ MNOCProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ UsableRulesProof
                                      ↳ QDP
                                        ↳ QReductionProof
QDP
                                            ↳ NonTerminationProof
                      ↳ QDP

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

ISPALISNEPAL
U711(tt) → ISPAL
ISNEPALU711(tt)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

ISPALISNEPAL
U711(tt) → ISPAL
ISNEPALU711(tt)

The TRS R consists of the following rules:none


s = ISNEPAL evaluates to t =ISNEPAL

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:




Rewriting sequence

ISNEPALU711(tt)
with rule ISNEPALU711(tt) at position [] and matcher [ ]

U711(tt)ISPAL
with rule U711(tt) → ISPAL at position [] and matcher [ ]

ISPALISNEPAL
with rule ISPALISNEPAL

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.





↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
QDP
                        ↳ UsableRulesProof

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

ISNELISTISNELIST
ISLISTISLIST
ISNELISTISLIST
ISLISTISNELIST
ISNELISTU511(isNeList)
U511(tt) → ISLIST
U411(tt) → ISNELIST
ISNELISTU411(isList)
ISLISTU211(isList)
U211(tt) → ISLIST

The TRS R consists of the following rules:

U11(tt) → tt
U21(tt) → U22(isList)
U22(tt) → tt
U31(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U71(tt) → U72(isPal)
U72(tt) → tt
U81(tt) → tt
isListU11(isNeList)
isListtt
isListU21(isList)
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
isNePalU71(isQid)
isPalU81(isNePal)
isQidtt

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
QDP
                            ↳ Narrowing

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

ISNELISTISLIST
ISLISTISLIST
ISNELISTISNELIST
ISLISTISNELIST
ISNELISTU511(isNeList)
U511(tt) → ISLIST
ISNELISTU411(isList)
U411(tt) → ISNELIST
U211(tt) → ISLIST
ISLISTU211(isList)

The TRS R consists of the following rules:

isListU11(isNeList)
isListtt
isListU21(isList)
U21(tt) → U22(isList)
U22(tt) → tt
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
U11(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
isQidtt
U31(tt) → tt

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule ISNELISTU511(isNeList) at position [0] we obtained the following new rules:

ISNELISTU511(U51(isNeList))
ISNELISTU511(U41(isList))
ISNELISTU511(U31(isQid))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ Narrowing
QDP
                                ↳ Narrowing

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

ISNELISTU511(U51(isNeList))
ISNELISTISLIST
ISLISTISLIST
ISNELISTISNELIST
ISLISTISNELIST
ISNELISTU511(U41(isList))
ISNELISTU411(isList)
U211(tt) → ISLIST
ISNELISTU511(U31(isQid))
U511(tt) → ISLIST
U411(tt) → ISNELIST
ISLISTU211(isList)

The TRS R consists of the following rules:

isListU11(isNeList)
isListtt
isListU21(isList)
U21(tt) → U22(isList)
U22(tt) → tt
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
U11(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
isQidtt
U31(tt) → tt

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule ISNELISTU411(isList) at position [0] we obtained the following new rules:

ISNELISTU411(U21(isList))
ISNELISTU411(tt)
ISNELISTU411(U11(isNeList))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Narrowing
QDP
                                    ↳ Narrowing

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

ISNELISTU511(U51(isNeList))
ISNELISTISNELIST
ISLISTISLIST
ISNELISTISLIST
ISNELISTU411(U11(isNeList))
ISLISTISNELIST
ISNELISTU511(U41(isList))
U211(tt) → ISLIST
ISNELISTU411(U21(isList))
ISNELISTU511(U31(isQid))
U511(tt) → ISLIST
ISNELISTU411(tt)
U411(tt) → ISNELIST
ISLISTU211(isList)

The TRS R consists of the following rules:

isListU11(isNeList)
isListtt
isListU21(isList)
U21(tt) → U22(isList)
U22(tt) → tt
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
U11(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
isQidtt
U31(tt) → tt

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule ISLISTU211(isList) at position [0] we obtained the following new rules:

ISLISTU211(U21(isList))
ISLISTU211(tt)
ISLISTU211(U11(isNeList))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ Narrowing
QDP
                                        ↳ Narrowing

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

ISNELISTU511(U51(isNeList))
ISNELISTISLIST
ISLISTISLIST
ISNELISTISNELIST
ISLISTISNELIST
ISNELISTU411(U11(isNeList))
ISLISTU211(U11(isNeList))
ISNELISTU511(U41(isList))
ISLISTU211(U21(isList))
U211(tt) → ISLIST
ISNELISTU411(U21(isList))
ISNELISTU511(U31(isQid))
U511(tt) → ISLIST
U411(tt) → ISNELIST
ISNELISTU411(tt)
ISLISTU211(tt)

The TRS R consists of the following rules:

isListU11(isNeList)
isListtt
isListU21(isList)
U21(tt) → U22(isList)
U22(tt) → tt
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
U11(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
isQidtt
U31(tt) → tt

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule ISNELISTU511(U31(isQid)) at position [0] we obtained the following new rules:

ISNELISTU511(U31(tt))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ Narrowing
                                      ↳ QDP
                                        ↳ Narrowing
QDP
                                            ↳ Narrowing

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

ISNELISTU511(U31(tt))
ISNELISTU511(U51(isNeList))
ISNELISTISNELIST
ISLISTISLIST
ISNELISTISLIST
ISNELISTU411(U11(isNeList))
ISLISTISNELIST
ISLISTU211(U11(isNeList))
ISNELISTU511(U41(isList))
ISLISTU211(U21(isList))
U211(tt) → ISLIST
ISNELISTU411(U21(isList))
U511(tt) → ISLIST
ISNELISTU411(tt)
U411(tt) → ISNELIST
ISLISTU211(tt)

The TRS R consists of the following rules:

isListU11(isNeList)
isListtt
isListU21(isList)
U21(tt) → U22(isList)
U22(tt) → tt
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
U11(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
isQidtt
U31(tt) → tt

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule ISNELISTU511(U31(tt)) at position [0] we obtained the following new rules:

ISNELISTU511(tt)



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ AND
                      ↳ QDP
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ Narrowing
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ Narrowing
QDP
                                                ↳ NonTerminationProof

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

ISNELISTU511(U51(isNeList))
ISNELISTISLIST
ISLISTISLIST
ISNELISTISNELIST
ISLISTISNELIST
ISNELISTU411(U11(isNeList))
ISLISTU211(U11(isNeList))
ISNELISTU511(U41(isList))
ISLISTU211(U21(isList))
U211(tt) → ISLIST
ISNELISTU411(U21(isList))
ISNELISTU511(tt)
U511(tt) → ISLIST
U411(tt) → ISNELIST
ISNELISTU411(tt)
ISLISTU211(tt)

The TRS R consists of the following rules:

isListU11(isNeList)
isListtt
isListU21(isList)
U21(tt) → U22(isList)
U22(tt) → tt
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
U11(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
isQidtt
U31(tt) → tt

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

ISNELISTU511(U51(isNeList))
ISNELISTISLIST
ISLISTISLIST
ISNELISTISNELIST
ISLISTISNELIST
ISNELISTU411(U11(isNeList))
ISLISTU211(U11(isNeList))
ISNELISTU511(U41(isList))
ISLISTU211(U21(isList))
U211(tt) → ISLIST
ISNELISTU411(U21(isList))
ISNELISTU511(tt)
U511(tt) → ISLIST
U411(tt) → ISNELIST
ISNELISTU411(tt)
ISLISTU211(tt)

The TRS R consists of the following rules:

isListU11(isNeList)
isListtt
isListU21(isList)
U21(tt) → U22(isList)
U22(tt) → tt
isNeListU31(isQid)
isNeListU41(isList)
isNeListU51(isNeList)
U11(tt) → tt
U51(tt) → U52(isList)
U52(tt) → tt
U41(tt) → U42(isNeList)
U42(tt) → tt
isQidtt
U31(tt) → tt


s = ISLIST evaluates to t =ISLIST

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:




Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from ISLIST to ISLIST.