YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 10 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) MRRProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) PisEmptyProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QReductionProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 1 ms]
        (22) YES


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

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

   a(b(x1)) -> b(b(a(x1)))
   c(b(x1)) -> b(c(c(x1)))

Q is empty.

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

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

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

   a(b(x1)) -> b(b(a(x1)))
   c(b(x1)) -> b(c(c(x1)))

The set Q consists of the following terms:

   a(b(x0))
   c(b(x0))


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

(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:

   A(b(x1)) -> A(x1)
   C(b(x1)) -> C(c(x1))
   C(b(x1)) -> C(x1)

The TRS R consists of the following rules:

   a(b(x1)) -> b(b(a(x1)))
   c(b(x1)) -> b(c(c(x1)))

The set Q consists of the following terms:

   a(b(x0))
   c(b(x0))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
----------------------------------------

(6)
Complex Obligation (AND)

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

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(b(x1)) -> C(x1)
   C(b(x1)) -> C(c(x1))

The TRS R consists of the following rules:

   a(b(x1)) -> b(b(a(x1)))
   c(b(x1)) -> b(c(c(x1)))

The set Q consists of the following terms:

   a(b(x0))
   c(b(x0))

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

(8) 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.
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(b(x1)) -> C(x1)
   C(b(x1)) -> C(c(x1))

The TRS R consists of the following rules:

   c(b(x1)) -> b(c(c(x1)))

The set Q consists of the following terms:

   a(b(x0))
   c(b(x0))

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

(10) 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].

   a(b(x0))


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(b(x1)) -> C(x1)
   C(b(x1)) -> C(c(x1))

The TRS R consists of the following rules:

   c(b(x1)) -> b(c(c(x1)))

The set Q consists of the following terms:

   c(b(x0))

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

(12) MRRProof (EQUIVALENT)
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:

   C(b(x1)) -> C(x1)
   C(b(x1)) -> C(c(x1))


Used ordering: Polynomial interpretation [POLO]:

   POL(C(x_1)) = 3*x_1
   POL(b(x_1)) = 1 + x_1
   POL(c(x_1)) = x_1


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

(13)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   c(b(x1)) -> b(c(c(x1)))

The set Q consists of the following terms:

   c(b(x0))

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

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

(15)
YES

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

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(b(x1)) -> A(x1)

The TRS R consists of the following rules:

   a(b(x1)) -> b(b(a(x1)))
   c(b(x1)) -> b(c(c(x1)))

The set Q consists of the following terms:

   a(b(x0))
   c(b(x0))

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

(17) 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.
----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(b(x1)) -> A(x1)

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

   a(b(x0))
   c(b(x0))

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

(19) 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].

   a(b(x0))
   c(b(x0))


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

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(b(x1)) -> A(x1)

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

(21) QDPSizeChangeProof (EQUIVALENT)
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:
*A(b(x1)) -> A(x1)
The graph contains the following edges 1 > 1


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(22)
YES