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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
  ↳3 QTRSRRRProof (⇔)
    ↳4 QTRS
    ↳5 AAECC Innermost (⇔)
    ↳6 QTRS
    ↳7 DependencyPairsProof (⇔)
    ↳8 QDP
    ↳9 DependencyGraphProof (⇔)
    ↳10 TRUE
  ↳11 QTRSRRRProof (⇔)
    ↳12 QTRS

(0) Obligation:

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

a__f(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a) = 0   
POL(a__b) = 0   
POL(a__f(x1, x2)) = 1 + 2·x1 + x2   
POL(b) = 0   
POL(f(x1, x2)) = 1 + 2·x1 + x2   
POL(mark(x1)) = 2·x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

mark(f(X1, X2)) → a__f(mark(X1), X2)


(2) Obligation:

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

a__f(X, X) → a__f(a, b)
a__b → a
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a) = 0   
POL(a__b) = 1   
POL(a__f(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(b) = 0   
POL(f(x1, x2)) = 1 + x1 + x2   
POL(mark(x1)) = 2 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__b → a
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b


(4) Obligation:

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

a__f(X, X) → a__f(a, b)

Q is empty.

(5) AAECC Innermost (EQUIVALENT transformation)

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

The TRS R 2 is

a__f(X, X) → a__f(a, b)

The signature Sigma is {a__f}

(6) Obligation:

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

a__f(X, X) → a__f(a, b)

The set Q consists of the following terms:

a__f(x0, x0)

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) Obligation:

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

A__F(X, X) → A__F(a, b)

The TRS R consists of the following rules:

a__f(X, X) → a__f(a, b)

The set Q consists of the following terms:

a__f(x0, x0)

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(10) TRUE

(11) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:
af2 > f2 > [a, b]
[ab, mark1] > [a, b]

Status:
a: multiset
ab: multiset
f2: multiset
af2: [1,2]
mark1: [1]
b: multiset

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

a__b → a
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b


(12) Obligation:

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

a__f(X, X) → a__f(a, b)

Q is empty.