YES Termination proof of ../tpdb/TRS/CSR/ExProp7_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(0) → cons(0, f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(X)) → X

The replacement map contains the following entries:

f: {1}
0: empty set
cons: {1}
s: {1}
p: {1}


↳ CSR
  ↳ Lucas-Transformation

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

f(0) → cons(0, f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(X)) → X

The replacement map contains the following entries:

f: {1}
0: empty set
cons: {1}
s: {1}
p: {1}

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

↳ CSR
  ↳ Lucas-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(X)) → X

Q is empty.

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

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(X)) → X

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

f(0) → cons(0)
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(cons(x1)) = 2·x1   
POL(f(x1)) = 2 + x1   
POL(p(x1)) = x1   
POL(s(x1)) = 2·x1   




↳ CSR
  ↳ Lucas-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof

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

f(s(0)) → f(p(s(0)))
p(s(X)) → X

Q is empty.

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

f(s(0)) → f(p(s(0)))
p(s(X)) → X

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

p(s(X)) → X
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(f(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + 2·x1   




↳ CSR
  ↳ Lucas-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ AAECC Innermost

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

f(s(0)) → f(p(s(0)))

Q is empty.

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

The TRS R 2 is

f(s(0)) → f(p(s(0)))

The signature Sigma is {f}

↳ CSR
  ↳ Lucas-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ AAECC Innermost
                ↳ QTRS
                  ↳ DependencyPairsProof

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

f(s(0)) → f(p(s(0)))

The set Q consists of the following terms:

f(s(0))


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(s(0)) → F(p(s(0)))

The TRS R consists of the following rules:

f(s(0)) → f(p(s(0)))

The set Q consists of the following terms:

f(s(0))

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

↳ CSR
  ↳ Lucas-Transformation
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ AAECC Innermost
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof

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

F(s(0)) → F(p(s(0)))

The TRS R consists of the following rules:

f(s(0)) → f(p(s(0)))

The set Q consists of the following terms:

f(s(0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.