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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 24 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 47 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 30 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) TRUE


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

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

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

Q is empty.

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

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

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

   B(b(c(x1))) -> A(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(A(x_1)) = x_1
   POL(B(x_1)) = x_1
   POL(C(x_1)) = 1 + x_1
   POL(a(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = 1 + x_1

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

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


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

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

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

The TRS R consists of the following rules:

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

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

(5) 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:

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

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = x_1
   POL(B(x_1)) = 2 + x_1
   POL(C(x_1)) = 2 + 3*x_1
   POL(a(x_1)) = 2 + x_1
   POL(b(x_1)) = 3 + x_1
   POL(c(x_1)) = 3 + 3*x_1


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

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

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

The TRS R consists of the following rules:

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

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

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
----------------------------------------

(8)
TRUE