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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 MNOCProof (⇔, 0 ms)
↳4 QDP
↳5 ForwardInstantiation (⇔, 0 ms)
↳6 QDP
↳7 MNOCProof (⇔, 0 ms)
↳8 QDP
↳9 NonTerminationLoopProof (⇔, 0 ms)
↳10 NO

(0) Obligation:

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

f(x, h(y)) → h(f(f(h(a), y), x))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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:

F(x, h(y)) → F(f(h(a), y), x)
F(x, h(y)) → F(h(a), y)

The TRS R consists of the following rules:

f(x, h(y)) → h(f(f(h(a), y), x))

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

(3) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(4) Obligation:

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

F(x, h(y)) → F(f(h(a), y), x)
F(x, h(y)) → F(h(a), y)

The TRS R consists of the following rules:

f(x, h(y)) → h(f(f(h(a), y), x))

The set Q consists of the following terms:

f(x0, h(x1))

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

(5) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(x, h(y)) → F(h(a), y) we obtained the following new rules [LPAR04]:

F(x0, h(h(y_1))) → F(h(a), h(y_1))

(6) Obligation:

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

F(x, h(y)) → F(f(h(a), y), x)
F(x0, h(h(y_1))) → F(h(a), h(y_1))

The TRS R consists of the following rules:

f(x, h(y)) → h(f(f(h(a), y), x))

The set Q consists of the following terms:

f(x0, h(x1))

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

(7) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(8) Obligation:

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

F(x, h(y)) → F(f(h(a), y), x)
F(x0, h(h(y_1))) → F(h(a), h(y_1))

The TRS R consists of the following rules:

f(x, h(y)) → h(f(f(h(a), y), x))

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

(9) NonTerminationLoopProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = F(x, f(x', f(x'', h(y)))) evaluates to t =F(f(h(a), f(f(h(a), f(f(h(a), y), x'')), x')), x)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [y / y', y' / y'', y'' / y''', x'' / x''', y''' / a, x''' / f(h(a), f(f(h(a), y), x''))]
  • Semiunifier: [x / f(h(y'''), f(x''', h(y'))), x' / h(y'')]



Rewriting sequence

F(f(h(y'''), f(x''', h(y'))), f(h(y''), f(x'', h(y)))) → F(f(h(y'''), f(x''', h(y'))), f(h(y''), h(f(f(h(a), y), x''))))
with rule f(x'''', h(y1)) → h(f(f(h(a), y1), x'''')) at position [1,1] and matcher [x'''' / x'', y1 / y]

F(f(h(y'''), f(x''', h(y'))), f(h(y''), h(f(f(h(a), y), x'')))) → F(f(h(y'''), f(x''', h(y'))), h(f(f(h(a), f(f(h(a), y), x'')), h(y''))))
with rule f(x', h(y'1)) → h(f(f(h(a), y'1), x')) at position [1] and matcher [x' / h(y''), y'1 / f(f(h(a), y), x'')]

F(f(h(y'''), f(x''', h(y'))), h(f(f(h(a), f(f(h(a), y), x'')), h(y'')))) → F(f(h(a), f(f(h(a), f(f(h(a), y), x'')), h(y''))), f(h(y'''), f(x''', h(y'))))
with rule F(x, h(y)) → F(f(h(a), y), x)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(10) NO