Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__f(x₁)) = x₁   
POL(a__p(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₁ + x₂   
POL(f(x₁)) = x₁   
POL(mark(x₁)) = 1 + x₁   
POL(p(x₁)) = x₁   
POL(s(x₁)) = x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

mark(0) → 0

(2) Obligation:

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__f(x₁)) = 2 + 2·x₁   
POL(a__p(x₁)) = 2·x₁   
POL(cons(x₁, x₂)) = x₁ + 2·x₂   
POL(f(x₁)) = 1 + 2·x₁   
POL(mark(x₁)) = 2·x₁   
POL(p(x₁)) = 2·x₁   
POL(s(x₁)) = 2·x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__f(X) → f(X)

(4) Obligation:

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__p(X) → p(X)

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__f(x₁)) = 2 + 2·x₁   
POL(a__p(x₁)) = 2·x₁   
POL(cons(x₁, x₂)) = 2·x₁ + x₂   
POL(f(x₁)) = 1 + 2·x₁   
POL(mark(x₁)) = 2·x₁   
POL(p(x₁)) = 2·x₁   
POL(s(x₁)) = x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__f(0) → cons(0, f(s(0)))

(6) Obligation:

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

a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__p(X) → p(X)

Q is empty.

(7) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__f(x₁)) = x₁   
POL(a__p(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₁ + x₂   
POL(f(x₁)) = 1 + x₁   
POL(mark(x₁)) = x₁   
POL(p(x₁)) = x₁   
POL(s(x₁)) = 1 + x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__p(s(0)) → 0
mark(f(X)) → a__f(mark(X))

(8) Obligation:

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

a__f(s(0)) → a__f(a__p(s(0)))
mark(p(X)) → a__p(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__p(X) → p(X)

Q is empty.

(9) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__f(x₁)) = x₁   
POL(a__p(x₁)) = 2·x₁   
POL(cons(x₁, x₂)) = 1 + x₁ + 2·x₂   
POL(mark(x₁)) = 2·x₁   
POL(p(x₁)) = 2·x₁   
POL(s(x₁)) = 2·x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

mark(cons(X1, X2)) → cons(mark(X1), X2)

(10) Obligation:

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

a__f(s(0)) → a__f(a__p(s(0)))
mark(p(X)) → a__p(mark(X))
mark(s(X)) → s(mark(X))
a__p(X) → p(X)

Q is empty.

(11) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__f(x₁)) = 2·x₁   
POL(a__p(x₁)) = x₁   
POL(mark(x₁)) = 2·x₁   
POL(p(x₁)) = x₁   
POL(s(x₁)) = 1 + x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

mark(s(X)) → s(mark(X))

(12) Obligation:

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

a__f(s(0)) → a__f(a__p(s(0)))
mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

Q is empty.

(13) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

The TRS R 2 is

a__f(s(0)) → a__f(a__p(s(0)))

The signature Sigma is {a__f}

(14) Obligation:

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

a__f(s(0)) → a__f(a__p(s(0)))
mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

The set Q consists of the following terms:

a__f(s(0))
mark(p(x0))
a__p(x0)

(15) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(16) Obligation:

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

A__F(s(0)) → A__F(a__p(s(0)))
A__F(s(0)) → A__P(s(0))
MARK(p(X)) → A__P(mark(X))
MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(s(0)) → a__f(a__p(s(0)))
mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

The set Q consists of the following terms:

a__f(s(0))
mark(p(x0))
a__p(x0)

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(18) Obligation:

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

MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(s(0)) → a__f(a__p(s(0)))
mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

The set Q consists of the following terms:

a__f(s(0))
mark(p(x0))
a__p(x0)

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

(19) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(20) Obligation:

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

MARK(p(X)) → MARK(X)

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

a__f(s(0))
mark(p(x0))
a__p(x0)

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

(21) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

a__f(s(0))
mark(p(x0))
a__p(x0)

(22) Obligation:

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

MARK(p(X)) → MARK(X)

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

(23) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• MARK(p(X)) → MARK(X)
  The graph contains the following edges 1 > 1

(24) TRUE