YES Termination proof of ../tpdb/TRS/CSR/Ex1_Zan97.trs
Termination of the following Term Rewriting System could be proven:

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

g(X) → h(X)
c → d
h(d) → g(c)

The replacement map contains the following entries:

g: empty set
h: empty set
c: empty set
d: empty set


↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation

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

g(X) → h(X)
c → d
h(d) → g(c)

The replacement map contains the following entries:

g: empty set
h: empty set
c: empty set
d: empty set

We applied the Lucas [26] to transform the context-sensitive TRS to an usual TRS.

↳ CSR
  ↳ Lucas-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation

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

g → h
c → d
h → g

Q is empty.

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

g → h
c → d
h → g

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

c → d
Used ordering:
Polynomial interpretation [25]:

POL(c) = 2   
POL(d) = 1   
POL(g) = 0   
POL(h) = 0   




↳ CSR
  ↳ Lucas-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation

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

g → h
h → g

Q is empty.

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

The TRS R 2 is

g → h
h → g

The signature Sigma is {g, h}

↳ CSR
  ↳ Lucas-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation

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

g → h
h → g

The set Q consists of the following terms:

g
h


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

H → G
G → H

The TRS R consists of the following rules:

g → h
h → g

The set Q consists of the following terms:

g
h

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

↳ CSR
  ↳ Lucas-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation

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

H → G
G → H

The TRS R consists of the following rules:

g → h
h → g

The set Q consists of the following terms:

g
h

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.

↳ CSR
  ↳ Lucas-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation

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

H → G
G → H

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

g
h

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.

g
h



↳ CSR
  ↳ Lucas-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof
                        ↳ QDP
                          ↳ NonTerminationProof
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation

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

H → G
G → H

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:

H → G
G → H

The TRS R consists of the following rules:none


s = G evaluates to t =G

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



Rewriting sequence

G → H
with rule G → H at position [] and matcher [ ]

H → G
with rule H → G

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.




We applied the Zantema [34] to transform the context-sensitive TRS to an usual TRS.

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

g(X) → h(a(X))
c → d
h(dInact) → g(cInact)
a(x) → x
d → dInact
a(dInact) → d
c → cInact
a(cInact) → c

Q is empty.

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

g(X) → h(a(X))
c → d
h(dInact) → g(cInact)
a(x) → x
d → dInact
a(dInact) → d
c → cInact
a(cInact) → c

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

a(x) → x
a(dInact) → d
c → cInact
Used ordering:
Polynomial interpretation [25]:

POL(a(x1)) = 1 + x1   
POL(c) = 1   
POL(cInact) = 0   
POL(d) = 1   
POL(dInact) = 1   
POL(g(x1)) = 1 + x1   
POL(h(x1)) = x1   




↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
  ↳ Incomplete Giesl Middeldorp-Transformation

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

g(X) → h(a(X))
c → d
h(dInact) → g(cInact)
d → dInact
a(cInact) → c

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

d → dInact
a(cInact) → c
c → d

The TRS R 2 is

g(X) → h(a(X))
h(dInact) → g(cInact)

The signature Sigma is {h, g}

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

g(X) → h(a(X))
c → d
h(dInact) → g(cInact)
d → dInact
a(cInact) → c

The set Q consists of the following terms:

g(x0)
c
h(dInact)
d
a(cInact)


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

A(cInact) → C
H(dInact) → G(cInact)
G(X) → H(a(X))
C → D
G(X) → A(X)

The TRS R consists of the following rules:

g(X) → h(a(X))
c → d
h(dInact) → g(cInact)
d → dInact
a(cInact) → c

The set Q consists of the following terms:

g(x0)
c
h(dInact)
d
a(cInact)

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

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

A(cInact) → C
H(dInact) → G(cInact)
G(X) → H(a(X))
C → D
G(X) → A(X)

The TRS R consists of the following rules:

g(X) → h(a(X))
c → d
h(dInact) → g(cInact)
d → dInact
a(cInact) → c

The set Q consists of the following terms:

g(x0)
c
h(dInact)
d
a(cInact)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(X) → H(a(X))
H(dInact) → G(cInact)

The TRS R consists of the following rules:

g(X) → h(a(X))
c → d
h(dInact) → g(cInact)
d → dInact
a(cInact) → c

The set Q consists of the following terms:

g(x0)
c
h(dInact)
d
a(cInact)

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.

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(X) → H(a(X))
H(dInact) → G(cInact)

The TRS R consists of the following rules:

a(cInact) → c
c → d
d → dInact

The set Q consists of the following terms:

g(x0)
c
h(dInact)
d
a(cInact)

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.

g(x0)
h(dInact)



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ Instantiation
  ↳ Incomplete Giesl Middeldorp-Transformation

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

H(dInact) → G(cInact)
G(X) → H(a(X))

The TRS R consists of the following rules:

a(cInact) → c
c → d
d → dInact

The set Q consists of the following terms:

c
d
a(cInact)

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule G(X) → H(a(X)) we obtained the following new rules:

G(cInact) → H(a(cInact))



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ Instantiation
                                ↳ QDP
                                  ↳ Rewriting
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(cInact) → H(a(cInact))
H(dInact) → G(cInact)

The TRS R consists of the following rules:

a(cInact) → c
c → d
d → dInact

The set Q consists of the following terms:

c
d
a(cInact)

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

G(cInact) → H(c)



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ Instantiation
                                ↳ QDP
                                  ↳ Rewriting
                                    ↳ QDP
                                      ↳ UsableRulesProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(cInact) → H(c)
H(dInact) → G(cInact)

The TRS R consists of the following rules:

a(cInact) → c
c → d
d → dInact

The set Q consists of the following terms:

c
d
a(cInact)

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.

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ Instantiation
                                ↳ QDP
                                  ↳ Rewriting
                                    ↳ QDP
                                      ↳ UsableRulesProof
                                        ↳ QDP
                                          ↳ QReductionProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(cInact) → H(c)
H(dInact) → G(cInact)

The TRS R consists of the following rules:

c → d
d → dInact

The set Q consists of the following terms:

c
d
a(cInact)

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.

a(cInact)



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ Instantiation
                                ↳ QDP
                                  ↳ Rewriting
                                    ↳ QDP
                                      ↳ UsableRulesProof
                                        ↳ QDP
                                          ↳ QReductionProof
                                            ↳ QDP
                                              ↳ Rewriting
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(cInact) → H(c)
H(dInact) → G(cInact)

The TRS R consists of the following rules:

c → d
d → dInact

The set Q consists of the following terms:

c
d

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

G(cInact) → H(d)



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ Instantiation
                                ↳ QDP
                                  ↳ Rewriting
                                    ↳ QDP
                                      ↳ UsableRulesProof
                                        ↳ QDP
                                          ↳ QReductionProof
                                            ↳ QDP
                                              ↳ Rewriting
                                                ↳ QDP
                                                  ↳ UsableRulesProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(cInact) → H(d)
H(dInact) → G(cInact)

The TRS R consists of the following rules:

c → d
d → dInact

The set Q consists of the following terms:

c
d

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.

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ Instantiation
                                ↳ QDP
                                  ↳ Rewriting
                                    ↳ QDP
                                      ↳ UsableRulesProof
                                        ↳ QDP
                                          ↳ QReductionProof
                                            ↳ QDP
                                              ↳ Rewriting
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ QReductionProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(cInact) → H(d)
H(dInact) → G(cInact)

The TRS R consists of the following rules:

d → dInact

The set Q consists of the following terms:

c
d

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.

c



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ Instantiation
                                ↳ QDP
                                  ↳ Rewriting
                                    ↳ QDP
                                      ↳ UsableRulesProof
                                        ↳ QDP
                                          ↳ QReductionProof
                                            ↳ QDP
                                              ↳ Rewriting
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ QReductionProof
                                                        ↳ QDP
                                                          ↳ Rewriting
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(cInact) → H(d)
H(dInact) → G(cInact)

The TRS R consists of the following rules:

d → dInact

The set Q consists of the following terms:

d

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

G(cInact) → H(dInact)



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ Instantiation
                                ↳ QDP
                                  ↳ Rewriting
                                    ↳ QDP
                                      ↳ UsableRulesProof
                                        ↳ QDP
                                          ↳ QReductionProof
                                            ↳ QDP
                                              ↳ Rewriting
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ QReductionProof
                                                        ↳ QDP
                                                          ↳ Rewriting
                                                            ↳ QDP
                                                              ↳ UsableRulesProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(cInact) → H(dInact)
H(dInact) → G(cInact)

The TRS R consists of the following rules:

d → dInact

The set Q consists of the following terms:

d

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.

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ Instantiation
                                ↳ QDP
                                  ↳ Rewriting
                                    ↳ QDP
                                      ↳ UsableRulesProof
                                        ↳ QDP
                                          ↳ QReductionProof
                                            ↳ QDP
                                              ↳ Rewriting
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ QReductionProof
                                                        ↳ QDP
                                                          ↳ Rewriting
                                                            ↳ QDP
                                                              ↳ UsableRulesProof
                                                                ↳ QDP
                                                                  ↳ QReductionProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(cInact) → H(dInact)
H(dInact) → G(cInact)

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

d

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.

d



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ Instantiation
                                ↳ QDP
                                  ↳ Rewriting
                                    ↳ QDP
                                      ↳ UsableRulesProof
                                        ↳ QDP
                                          ↳ QReductionProof
                                            ↳ QDP
                                              ↳ Rewriting
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ QReductionProof
                                                        ↳ QDP
                                                          ↳ Rewriting
                                                            ↳ QDP
                                                              ↳ UsableRulesProof
                                                                ↳ QDP
                                                                  ↳ QReductionProof
                                                                    ↳ QDP
                                                                      ↳ NonTerminationProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

G(cInact) → H(dInact)
H(dInact) → G(cInact)

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:

G(cInact) → H(dInact)
H(dInact) → G(cInact)

The TRS R consists of the following rules:none


s = H(dInact) evaluates to t =H(dInact)

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



Rewriting sequence

H(dInact) → G(cInact)
with rule H(dInact) → G(cInact) at position [] and matcher [ ]

G(cInact) → H(dInact)
with rule G(cInact) → H(dInact)

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.




We applied the Incomplete Giesl Middeldorp [11] to transform the context-sensitive TRS to an usual TRS.

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

mark(g(x1)) → gActive(x1)
gActive(x1) → g(x1)
mark(c) → cActive
cActive → c
mark(h(x1)) → hActive(x1)
hActive(x1) → h(x1)
mark(d) → d
gActive(X) → hActive(X)
cActive → d
hActive(d) → gActive(c)

Q is empty.

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

mark(g(x1)) → gActive(x1)
gActive(x1) → g(x1)
mark(c) → cActive
cActive → c
mark(h(x1)) → hActive(x1)
hActive(x1) → h(x1)
mark(d) → d
gActive(X) → hActive(X)
cActive → d
hActive(d) → gActive(c)

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

gActive(x1) → g(x1)
mark(c) → cActive
cActive → c
mark(h(x1)) → hActive(x1)
mark(d) → d
gActive(X) → hActive(X)
hActive(d) → gActive(c)
Used ordering:
Polynomial interpretation [25]:

POL(c) = 1   
POL(cActive) = 2   
POL(d) = 2   
POL(g(x1)) = x1   
POL(gActive(x1)) = 1 + 2·x1   
POL(h(x1)) = 2·x1   
POL(hActive(x1)) = 2·x1   
POL(mark(x1)) = 1 + 2·x1   




↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof

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

mark(g(x1)) → gActive(x1)
hActive(x1) → h(x1)
cActive → d

Q is empty.

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

mark(g(x1)) → gActive(x1)
hActive(x1) → h(x1)
cActive → d

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

mark(g(x1)) → gActive(x1)
hActive(x1) → h(x1)
cActive → d
Used ordering:
Polynomial interpretation [25]:

POL(cActive) = 2   
POL(d) = 1   
POL(g(x1)) = 2 + 2·x1   
POL(gActive(x1)) = 1 + x1   
POL(h(x1)) = 1 + x1   
POL(hActive(x1)) = 2 + 2·x1   
POL(mark(x1)) = 2 + 2·x1   




↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
  ↳ Incomplete Giesl Middeldorp-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ RisEmptyProof

Q restricted rewrite system:
R is empty.
Q is empty.

The TRS R is empty. Hence, termination is trivially proven.