YES Termination w.r.t. Q proof of Transformed_CSR_04_ExProp7_Luc06_GM.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 6 ms)
↳4 QTRS
↳5 QTRSRRRProof (⇔, 0 ms)
↳6 QTRS
↳7 QTRSRRRProof (⇔, 0 ms)
↳8 QTRS
↳9 QTRSRRRProof (⇔, 0 ms)
↳10 QTRS
↳11 QTRSRRRProof (⇔, 6 ms)
↳12 QTRS
↳13 QTRSRRRProof (⇔, 0 ms)
↳14 QTRS
↳15 AAECC Innermost (⇔, 0 ms)
↳16 QTRS
↳17 DependencyPairsProof (⇔, 0 ms)
↳18 QDP
↳19 DependencyGraphProof (⇔, 0 ms)
↳20 QDP
↳21 UsableRulesProof (⇔, 0 ms)
↳22 QDP
↳23 QReductionProof (⇔, 0 ms)
↳24 QDP
↳25 QDPSizeChangeProof (⇔, 0 ms)
↳26 YES

(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(X)) → mark(X)
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)) = 2 + 2·x1   
POL(a__p(x1)) = x1   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(f(x1)) = 1 + 2·x1   
POL(mark(x1)) = 2·x1   
POL(p(x1)) = 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)


(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(X)) → mark(X)
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__p(X) → p(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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

mark(f(X)) → a__f(mark(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(X)) → 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__p(X) → p(X)

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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


(6) 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(X)) → mark(X)
mark(p(X)) → a__p(mark(X))
mark(0) → 0
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) = 2   
POL(a__f(x1)) = 2 + 2·x1   
POL(a__p(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(f(x1)) = x1   
POL(mark(x1)) = 2·x1   
POL(p(x1)) = 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(0) → 0


(8) 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(X)) → mark(X)
mark(p(X)) → a__p(mark(X))
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)) = 2 + 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 + 2·x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__p(s(X)) → mark(X)


(10) 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)))
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) = 2   
POL(a__f(x1)) = 2 + 2·x1   
POL(a__p(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(f(x1)) = x1   
POL(mark(x1)) = 2 + 2·x1   
POL(p(x1)) = 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(s(X)) → s(mark(X))


(12) 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)))
mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

Q is empty.

(13) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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

a__f(0) → cons(0, f(s(0)))


(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)

Q is empty.

(15) 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}

(16) 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)

(17) DependencyPairsProof (EQUIVALENT transformation)

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

(18) 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.

(19) DependencyGraphProof (EQUIVALENT transformation)

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

(20) 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.

(21) 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.

(22) 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.

(23) 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)

(24) 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.

(25) 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

(26) YES