YES Termination proof of ../tpdb/TRS/CSR/Ex9_Luc06.trs
Termination of the following Term Rewriting System could be proven:

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

f(a, X, X) → f(X, b, b)
b → a

The replacement map contains the following entries:

f: {2}
a: empty set
b: empty set


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

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

f(a, X, X) → f(X, b, b)
b → a

The replacement map contains the following entries:

f: {2}
a: empty set
b: empty set

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

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

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

f(X) → f(b)
b → a

Q is empty.

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

b → a

The TRS R 2 is

f(X) → f(b)

The signature Sigma is {f}

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

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

f(X) → f(b)
b → a

The set Q consists of the following terms:

f(x0)
b


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

F(X) → B
F(X) → F(b)

The TRS R consists of the following rules:

f(X) → f(b)
b → a

The set Q consists of the following terms:

f(x0)
b

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

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

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

F(X) → B
F(X) → F(b)

The TRS R consists of the following rules:

f(X) → f(b)
b → a

The set Q consists of the following terms:

f(x0)
b

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

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

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

F(X) → F(b)

The TRS R consists of the following rules:

f(X) → f(b)
b → a

The set Q consists of the following terms:

f(x0)
b

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

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

F(X) → F(b)

The TRS R consists of the following rules:

b → a

The set Q consists of the following terms:

f(x0)
b

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.

f(x0)



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

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

F(X) → F(b)

The TRS R consists of the following rules:

b → a

The set Q consists of the following terms:

b

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

F(X) → F(a)



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

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

F(X) → F(a)

The TRS R consists of the following rules:

b → a

The set Q consists of the following terms:

b

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

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

F(X) → F(a)

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

b

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.

b



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

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

F(X) → F(a)

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

F(a) → F(a)



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

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

F(a) → F(a)

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:

F(a) → F(a)

The TRS R consists of the following rules:none


s = F(a) evaluates to t =F(a)

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 F(a) to F(a).




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

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

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

f(aInact, X, X) → f(a(X), b, bInact)
b → a
a(x) → x
b → bInact
a(bInact) → b
a → aInact
a(aInact) → a

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:

B → A
F(aInact, X, X) → A(X)
F(aInact, X, X) → F(a(X), b, bInact)
A(bInact) → B
F(aInact, X, X) → B
A(aInact) → A

The TRS R consists of the following rules:

f(aInact, X, X) → f(a(X), b, bInact)
b → a
a(x) → x
b → bInact
a(bInact) → b
a → aInact
a(aInact) → a

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

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

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

B → A
F(aInact, X, X) → A(X)
F(aInact, X, X) → F(a(X), b, bInact)
A(bInact) → B
F(aInact, X, X) → B
A(aInact) → A

The TRS R consists of the following rules:

f(aInact, X, X) → f(a(X), b, bInact)
b → a
a(x) → x
b → bInact
a(bInact) → b
a → aInact
a(aInact) → a

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 5 less nodes.

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

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

F(aInact, X, X) → F(a(X), b, bInact)

The TRS R consists of the following rules:

f(aInact, X, X) → f(a(X), b, bInact)
b → a
a(x) → x
b → bInact
a(bInact) → b
a → aInact
a(aInact) → a

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.

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, X, X) → F(a(X), b, bInact)

The TRS R consists of the following rules:

a(x) → x
a(bInact) → b
a(aInact) → a
b → a
b → bInact
a → aInact

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(aInact, X, X) → F(a(X), b, bInact) we obtained the following new rules:

F(aInact, bInact, bInact) → F(a(bInact), b, bInact)



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(a(bInact), b, bInact)

The TRS R consists of the following rules:

a(x) → x
a(bInact) → b
a(aInact) → a
b → a
b → bInact
a → aInact

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.

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(a(bInact), b, bInact)

The TRS R consists of the following rules:

a(x) → x
a(bInact) → b
b → a
b → bInact
a → aInact

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

F(aInact, bInact, bInact) → F(bInact, b, bInact)
F(aInact, bInact, bInact) → F(b, b, bInact)



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(bInact, b, bInact)
F(aInact, bInact, bInact) → F(b, b, bInact)

The TRS R consists of the following rules:

a(x) → x
a(bInact) → b
b → a
b → bInact
a → aInact

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

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(b, b, bInact)

The TRS R consists of the following rules:

a(x) → x
a(bInact) → b
b → a
b → bInact
a → aInact

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.

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(b, b, bInact)

The TRS R consists of the following rules:

b → a
b → bInact
a → aInact

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule F(aInact, bInact, bInact) → F(b, b, bInact) at position [1] we obtained the following new rules:

F(aInact, bInact, bInact) → F(b, bInact, bInact)
F(aInact, bInact, bInact) → F(b, a, bInact)



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(b, a, bInact)
F(aInact, bInact, bInact) → F(b, bInact, bInact)

The TRS R consists of the following rules:

b → a
b → bInact
a → aInact

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule F(aInact, bInact, bInact) → F(b, a, bInact) at position [1] we obtained the following new rules:

F(aInact, bInact, bInact) → F(b, aInact, bInact)



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ DependencyGraphProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(b, bInact, bInact)
F(aInact, bInact, bInact) → F(b, aInact, bInact)

The TRS R consists of the following rules:

b → a
b → bInact
a → aInact

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

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(b, bInact, bInact)

The TRS R consists of the following rules:

b → a
b → bInact
a → aInact

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

b → bInact

Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x1, x2, x3)) = 2·x1 + x2 + x3   
POL(a) = 1   
POL(aInact) = 1   
POL(b) = 1   
POL(bInact) = 0   



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ MNOCProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(b, bInact, bInact)

The TRS R consists of the following rules:

b → a
a → aInact

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.

↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ MNOCProof
                                                        ↳ QDP
                                                          ↳ Rewriting
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(b, bInact, bInact)

The TRS R consists of the following rules:

b → a
a → aInact

The set Q consists of the following terms:

b
a

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

F(aInact, bInact, bInact) → F(a, bInact, bInact)



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ MNOCProof
                                                        ↳ QDP
                                                          ↳ Rewriting
                                                            ↳ QDP
                                                              ↳ UsableRulesProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(a, bInact, bInact)

The TRS R consists of the following rules:

b → a
a → aInact

The set Q consists of the following terms:

b
a

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
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ MNOCProof
                                                        ↳ QDP
                                                          ↳ Rewriting
                                                            ↳ QDP
                                                              ↳ UsableRulesProof
                                                                ↳ QDP
                                                                  ↳ QReductionProof
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(a, bInact, bInact)

The TRS R consists of the following rules:

a → aInact

The set Q consists of the following terms:

b
a

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.

b



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ MNOCProof
                                                        ↳ QDP
                                                          ↳ Rewriting
                                                            ↳ QDP
                                                              ↳ UsableRulesProof
                                                                ↳ QDP
                                                                  ↳ QReductionProof
                                                                    ↳ QDP
                                                                      ↳ Rewriting
  ↳ Incomplete Giesl Middeldorp-Transformation

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

F(aInact, bInact, bInact) → F(a, bInact, bInact)

The TRS R consists of the following rules:

a → aInact

The set Q consists of the following terms:

a

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

F(aInact, bInact, bInact) → F(aInact, bInact, bInact)



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ MNOCProof
                                                        ↳ 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:

F(aInact, bInact, bInact) → F(aInact, bInact, bInact)

The TRS R consists of the following rules:

a → aInact

The set Q consists of the following terms:

a

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
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ MNOCProof
                                                        ↳ 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:

F(aInact, bInact, bInact) → F(aInact, bInact, bInact)

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

a

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



↳ CSR
  ↳ Lucas-Transformation
  ↳ Zantema-Transformation
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ UsableRulesProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ MNOCProof
                                                        ↳ 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:

F(aInact, bInact, bInact) → F(aInact, bInact, bInact)

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:

F(aInact, bInact, bInact) → F(aInact, bInact, bInact)

The TRS R consists of the following rules:none


s = F(aInact, bInact, bInact) evaluates to t =F(aInact, bInact, bInact)

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 F(aInact, bInact, bInact) to F(aInact, bInact, bInact).




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(f(x1, x2, x3)) → fActive(x1, mark(x2), x3)
fActive(x1, x2, x3) → f(x1, x2, x3)
mark(b) → bActive
bActive → b
mark(a) → a
fActive(a, X, X) → fActive(X, bActive, b)
bActive → a

Q is empty.

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

mark(f(x1, x2, x3)) → fActive(x1, mark(x2), x3)
fActive(x1, x2, x3) → f(x1, x2, x3)
mark(b) → bActive
bActive → b
mark(a) → a
fActive(a, X, X) → fActive(X, bActive, b)
bActive → a

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

fActive(x1, x2, x3) → f(x1, x2, x3)
bActive → b
mark(a) → a
bActive → a
Used ordering:
Polynomial interpretation [25]:

POL(a) = 1   
POL(b) = 0   
POL(bActive) = 2   
POL(f(x1, x2, x3)) = 1 + 2·x1 + x2 + 2·x3   
POL(fActive(x1, x2, x3)) = 2 + 2·x1 + x2 + 2·x3   
POL(mark(x1)) = 2 + 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(f(x1, x2, x3)) → fActive(x1, mark(x2), x3)
mark(b) → bActive
fActive(a, X, X) → fActive(X, bActive, b)

Q is empty.

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

mark(f(x1, x2, x3)) → fActive(x1, mark(x2), x3)
mark(b) → bActive
fActive(a, X, X) → fActive(X, bActive, b)

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(f(x1, x2, x3)) → fActive(x1, mark(x2), x3)
mark(b) → bActive
fActive(a, X, X) → fActive(X, bActive, b)
Used ordering:
Polynomial interpretation [25]:

POL(a) = 2   
POL(b) = 1   
POL(bActive) = 0   
POL(f(x1, x2, x3)) = 2 + 2·x1 + 2·x2 + 2·x3   
POL(fActive(x1, x2, x3)) = 2·x1 + x2 + x3   
POL(mark(x1)) = 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.