Termination of the following Term Rewriting System could be proven:

Context-sensitive 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
  ↳ CSRInnermostProof

Context-sensitive 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}

The CSR is orthogonal. By [10] we can switch to innermost.

↳ CSR
  ↳ CSRInnermostProof
    ↳ CSR
      ↳ CSDependencyPairsProof

Context-sensitive 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}

Innermost Strategy.

Using Improved CS-DPs we result in the following initial Q-CSDP problem.

↳ CSR
  ↳ CSRInnermostProof
    ↳ CSR
      ↳ CSDependencyPairsProof
        ↳ QCSDP
          ↳ QCSDependencyGraphProof

Q-restricted context-sensitive dependency pair problem:
The symbols in {f, s, p, F, P} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.
The symbols in {U} are not replacing on any position.

The ordinary context-sensitive dependency pairs DPo are:

F(s(0)) → F(p(s(0)))
F(s(0)) → P(s(0))


The hidden terms of R are:

f(s(0))

Every hiding context is built from:none

Hence, the new unhiding pairs DPu are :

U(f(s(0))) → F(s(0))

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 set Q consists of the following terms:

f(0)
f(s(0))
p(s(x0))


The approximation of the Context-Sensitive Dependency Graph contains 1 SCC with 2 less nodes.


↳ CSR
  ↳ CSRInnermostProof
    ↳ CSR
      ↳ CSDependencyPairsProof
        ↳ QCSDP
          ↳ QCSDependencyGraphProof
            ↳ QCSDP
              ↳ QCSUsableRulesProof

Q-restricted context-sensitive dependency pair problem:
The symbols in {f, s, p, F} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The TRS P consists of the following rules:

F(s(0)) → F(p(s(0)))

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 set Q consists of the following terms:

f(0)
f(s(0))
p(s(x0))


The following rules are not useable and can be deleted:

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


↳ CSR
  ↳ CSRInnermostProof
    ↳ CSR
      ↳ CSDependencyPairsProof
        ↳ QCSDP
          ↳ QCSDependencyGraphProof
            ↳ QCSDP
              ↳ QCSUsableRulesProof
                ↳ QCSDP
                  ↳ QCSDPReductionPairProof

Q-restricted context-sensitive dependency pair problem:
The symbols in {p, s, F, f} are replacing on all positions.

The TRS P consists of the following rules:

F(s(0)) → F(p(s(0)))

The TRS R consists of the following rules:

p(s(X)) → X

The set Q consists of the following terms:

f(0)
f(s(0))
p(s(x0))


Using the order
Polynomial interpretation [25,35]:

POL(s(x1)) = 4 + (3/2)x1   
POL(p(x1)) = (3/4)x1   
POL(0) = 4   
POL(F(x1)) = (1/4)x1   
The value of delta used in the strict ordering is 5/8.
the following usable rules

p(s(X)) → X

could all be oriented weakly.
Furthermore, the pairs

F(s(0)) → F(p(s(0)))

could be oriented strictly and thus removed.
All pairs have been removed.


↳ CSR
  ↳ CSRInnermostProof
    ↳ CSR
      ↳ CSDependencyPairsProof
        ↳ QCSDP
          ↳ QCSDependencyGraphProof
            ↳ QCSDP
              ↳ QCSUsableRulesProof
                ↳ QCSDP
                  ↳ QCSDPReductionPairProof
                    ↳ QCSDP
                      ↳ PIsEmptyProof

Q-restricted context-sensitive dependency pair problem:
The symbols in {p, s, f} are replacing on all positions.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

p(s(X)) → X

The set Q consists of the following terms:

f(0)
f(s(0))
p(s(x0))


The TRS P is empty. Hence, there is no (P,Q,R,µ)-chain.