YES Termination w.r.t. Q proof of Transformed_CSR_04_Ex25_Luc06_FR.ari

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 38 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 QDPOrderProof (⇔, 18 ms)
        ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 MRRProof (⇔, 10 ms)
        ↳25 QDP
        ↳26 PisEmptyProof (⇔, 0 ms)
        ↳27 YES

(0) Obligation:

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

f(f(X)) → c(n__f(n__g(n__f(X))))
c(X) → d(activate(X))
h(X) → c(n__d(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(X)
activate(n__d(X)) → d(X)
activate(X) → X

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

f(f(X)) → n__f(n__g(n__f(c(X))))
c(X) → activate(d(X))
h(X) → n__d(c(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
n__f(activate(X)) → activate(f(X))
n__g(activate(X)) → g(X)
n__d(activate(X)) → d(X)
activate(X) → X

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(activate(x1)) = x1   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
POL(f(x1)) = x1   
POL(g(x1)) = x1   
POL(h(x1)) = 1 + x1   
POL(n__d(x1)) = x1   
POL(n__f(x1)) = x1   
POL(n__g(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

h(X) → n__d(c(X))


(4) Obligation:

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

f(f(X)) → n__f(n__g(n__f(c(X))))
c(X) → activate(d(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
n__f(activate(X)) → activate(f(X))
n__g(activate(X)) → g(X)
n__d(activate(X)) → d(X)
activate(X) → X

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

F(f(X)) → N__F(n__g(n__f(c(X))))
F(f(X)) → N__G(n__f(c(X)))
F(f(X)) → N__F(c(X))
F(f(X)) → C(X)
C(X) → ACTIVATE(d(X))
C(X) → D(X)
F(X) → N__F(X)
G(X) → N__G(X)
D(X) → N__D(X)
N__F(activate(X)) → ACTIVATE(f(X))
N__F(activate(X)) → F(X)
N__G(activate(X)) → G(X)
N__D(activate(X)) → D(X)

The TRS R consists of the following rules:

f(f(X)) → n__f(n__g(n__f(c(X))))
c(X) → activate(d(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
n__f(activate(X)) → activate(f(X))
n__g(activate(X)) → g(X)
n__d(activate(X)) → d(X)
activate(X) → X

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 5 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

N__D(activate(X)) → D(X)
D(X) → N__D(X)

The TRS R consists of the following rules:

f(f(X)) → n__f(n__g(n__f(c(X))))
c(X) → activate(d(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
n__f(activate(X)) → activate(f(X))
n__g(activate(X)) → g(X)
n__d(activate(X)) → d(X)
activate(X) → X

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

(10) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(11) Obligation:

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

N__D(activate(X)) → D(X)
D(X) → N__D(X)

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

(12) 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:

  • D(X) → N__D(X)
    The graph contains the following edges 1 >= 1

  • N__D(activate(X)) → D(X)
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

N__G(activate(X)) → G(X)
G(X) → N__G(X)

The TRS R consists of the following rules:

f(f(X)) → n__f(n__g(n__f(c(X))))
c(X) → activate(d(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
n__f(activate(X)) → activate(f(X))
n__g(activate(X)) → g(X)
n__d(activate(X)) → d(X)
activate(X) → X

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

(15) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

N__G(activate(X)) → G(X)
G(X) → N__G(X)

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

(17) 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:

  • G(X) → N__G(X)
    The graph contains the following edges 1 >= 1

  • N__G(activate(X)) → G(X)
    The graph contains the following edges 1 > 1

(18) YES

(19) Obligation:

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

N__F(activate(X)) → F(X)
F(f(X)) → N__F(n__g(n__f(c(X))))
F(f(X)) → N__F(c(X))
F(X) → N__F(X)

The TRS R consists of the following rules:

f(f(X)) → n__f(n__g(n__f(c(X))))
c(X) → activate(d(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
n__f(activate(X)) → activate(f(X))
n__g(activate(X)) → g(X)
n__d(activate(X)) → d(X)
activate(X) → X

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


F(f(X)) → N__F(n__g(n__f(c(X))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(F(x1)) = 1 + x1   
POL(N__F(x1)) = 1 + x1   
POL(activate(x1)) = x1   
POL(c(x1)) = 1   
POL(d(x1)) = 0   
POL(f(x1)) = 1   
POL(g(x1)) = 0   
POL(n__d(x1)) = 0   
POL(n__f(x1)) = 1   
POL(n__g(x1)) = 0   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

c(X) → activate(d(X))
n__g(activate(X)) → g(X)
g(X) → n__g(X)
activate(X) → X
n__d(activate(X)) → d(X)
d(X) → n__d(X)

(21) Obligation:

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

N__F(activate(X)) → F(X)
F(f(X)) → N__F(c(X))
F(X) → N__F(X)

The TRS R consists of the following rules:

f(f(X)) → n__f(n__g(n__f(c(X))))
c(X) → activate(d(X))
f(X) → n__f(X)
g(X) → n__g(X)
d(X) → n__d(X)
n__f(activate(X)) → activate(f(X))
n__g(activate(X)) → g(X)
n__d(activate(X)) → d(X)
activate(X) → X

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

(22) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

N__F(activate(X)) → F(X)
F(f(X)) → N__F(c(X))
F(X) → N__F(X)

The TRS R consists of the following rules:

c(X) → activate(d(X))
activate(X) → X
n__d(activate(X)) → d(X)
d(X) → n__d(X)

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

(24) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

N__F(activate(X)) → F(X)
F(f(X)) → N__F(c(X))
F(X) → N__F(X)

Strictly oriented rules of the TRS R:

activate(X) → X
n__d(activate(X)) → d(X)

Used ordering: Polynomial interpretation [POLO]:

POL(F(x1)) = 2 + 3·x1   
POL(N__F(x1)) = 1 + 3·x1   
POL(activate(x1)) = 1 + x1   
POL(c(x1)) = 1 + 3·x1   
POL(d(x1)) = 2·x1   
POL(f(x1)) = 3 + 3·x1   
POL(n__d(x1)) = 2·x1   

(25) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

c(X) → activate(d(X))
d(X) → n__d(X)

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) YES