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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 2 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) TRUE


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

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

   a(b(a(b(x)))) -> b(a(b(a(a(b(x))))))

The set Q consists of the following terms:

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


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

(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(b(a(b(x)))) -> A(b(a(a(b(x)))))
   A(b(a(b(x)))) -> A(a(b(x)))

The TRS R consists of the following rules:

   a(b(a(b(x)))) -> b(a(b(a(a(b(x))))))

The set Q consists of the following terms:

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

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

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

(4)
TRUE