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:

eq → true
eq → eq
eq → false
inf(X) → cons
take(0, X) → nil
take(s, cons) → cons
length(nil) → 0
length(cons) → s

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

eq → true
eq → eq
eq → false
inf(X) → cons
take(0, X) → nil
take(s, cons) → cons
length(nil) → 0
length(cons) → s

Q is empty.

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

eq → true
eq → eq
eq → false
inf(X) → cons
take(0, X) → nil
take(s, cons) → cons
length(nil) → 0
length(cons) → s

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

eq → true
eq → false
take(s, cons) → cons
length(nil) → 0
length(cons) → s
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(cons) = 2   
POL(eq) = 2   
POL(false) = 0   
POL(inf(x1)) = 2 + x1   
POL(length(x1)) = 2 + 2·x1   
POL(nil) = 0   
POL(s) = 1   
POL(take(x1, x2)) = 2·x1 + 2·x2   
POL(true) = 1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

eq → eq
inf(X) → cons
take(0, X) → nil

Q is empty.

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

eq → eq
inf(X) → cons
take(0, X) → nil

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

inf(X) → cons
take(0, X) → nil
Used ordering:
Polynomial interpretation [25]:

POL(0) = 2   
POL(cons) = 1   
POL(eq) = 0   
POL(inf(x1)) = 2 + x1   
POL(nil) = 1   
POL(take(x1, x2)) = 2 + 2·x1 + x2   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost

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

eq → eq

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

eq → eq

The signature Sigma is {eq}

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof

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

eq → eq

The set Q consists of the following terms:

eq


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

EQ → EQ

The TRS R consists of the following rules:

eq → eq

The set Q consists of the following terms:

eq

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof

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

EQ → EQ

The TRS R consists of the following rules:

eq → eq

The set Q consists of the following terms:

eq

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
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof

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

EQ → EQ

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

eq

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.

eq



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof
                        ↳ QDP
                          ↳ NonTerminationProof

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

EQ → EQ

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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

EQ → EQ

The TRS R consists of the following rules:none


s = EQ evaluates to t =EQ

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 EQ to EQ.