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

(0) Obligation:

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

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

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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:

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

The set Q consists of the following terms:

f(h(x0), x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(h(x0), x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(h(x0), x1)

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

(7) ForwardInstantiation (EQUIVALENT transformation)

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

F(h(h(y_0)), x1) → F(h(y_0), h(f(a, a)))

(8) Obligation:

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

F(h(x), y) → F(y, f(x, h(f(a, a))))
F(h(h(y_0)), x1) → F(h(y_0), h(f(a, a)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(h(x0), x1)

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

(9) MNOCProof (EQUIVALENT transformation)

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

(10) Obligation:

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

F(h(x), y) → F(y, f(x, h(f(a, a))))
F(h(h(y_0)), x1) → F(h(y_0), h(f(a, a)))

The TRS R consists of the following rules:

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

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

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

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Matcher: [y'' / y''', x / x', x' / x'', x'' / x''', y''' / f(f(y'', f(x, h(f(a, a)))), h(f(a, a))), x''' / f(a, a)]
• Semiunifier: [y' / h(x''), y / f(f(h(x'), y'''), h(x'''))]

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Rewriting sequence

F(f(f(h(x), y''), h(x'')), f(f(h(x'), y'''), h(x'''))) → F(f(h(f(y'', f(x, h(f(a, a))))), h(x'')), f(f(h(x'), y'''), h(x''')))
with rule f(h(x1), y'''') → h(f(y'''', f(x1, h(f(a, a))))) at position [0,0] and matcher [x1 / x, y'''' / y'']

F(f(h(f(y'', f(x, h(f(a, a))))), h(x'')), f(f(h(x'), y'''), h(x'''))) → F(h(f(h(x''), f(f(y'', f(x, h(f(a, a)))), h(f(a, a))))), f(f(h(x'), y'''), h(x''')))
with rule f(h(x'1), y') → h(f(y', f(x'1, h(f(a, a))))) at position [0] and matcher [x'1 / f(y'', f(x, h(f(a, a)))), y' / h(x'')]

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

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.

(12) NO