YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QReductionProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QDPOrderProof [EQUIVALENT, 29 ms]
(12) QDP
(13) PisEmptyProof [EQUIVALENT, 0 ms]
(14) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(a, y) -> f(y, g(y))
   g(a) -> b
   g(b) -> b

Q is empty.

----------------------------------------

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   g(a) -> b
   g(b) -> b

The TRS R 2 is 
   f(a, y) -> f(y, g(y))

The signature Sigma is {f_2}
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(a, y) -> f(y, g(y))
   g(a) -> b
   g(b) -> b

The set Q consists of the following terms:

   f(a, x0)
   g(a)
   g(b)


----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a, y) -> F(y, g(y))
   F(a, y) -> G(y)

The TRS R consists of the following rules:

   f(a, y) -> f(y, g(y))
   g(a) -> b
   g(b) -> b

The set Q consists of the following terms:

   f(a, x0)
   g(a)
   g(b)

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a, y) -> F(y, g(y))

The TRS R consists of the following rules:

   f(a, y) -> f(y, g(y))
   g(a) -> b
   g(b) -> b

The set Q consists of the following terms:

   f(a, x0)
   g(a)
   g(b)

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

(7) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a, y) -> F(y, g(y))

The TRS R consists of the following rules:

   g(a) -> b
   g(b) -> b

The set Q consists of the following terms:

   f(a, x0)
   g(a)
   g(b)

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

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

   f(a, x0)


----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(a, y) -> F(y, g(y))

The TRS R consists of the following rules:

   g(a) -> b
   g(b) -> b

The set Q consists of the following terms:

   g(a)
   g(b)

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

(11) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   F(a, y) -> F(y, g(y))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
F(x1, x2)  =  F(x1, x2)

a  =  a

g(x1)  =  g

b  =  b


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   a=3
   b=1
   F_2=1
   g=2

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

   g(a) -> b
   g(b) -> b


----------------------------------------

(12)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   g(a) -> b
   g(b) -> b

The set Q consists of the following terms:

   g(a)
   g(b)

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

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

(14)
YES