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(x, x) → f(i(x), g(g(x)))
f(x, y) → x
g(x) → i(x)
f(x, i(x)) → f(x, x)
f(i(x), i(g(x))) → a

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a) = 1   
POL(f(x₁, x₂)) = 2 + 2·x₁ + 2·x₂   
POL(g(x₁)) = x₁   
POL(i(x₁)) = x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

f(x, y) → x
f(i(x), i(g(x))) → a

(2) Obligation:

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

f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)

Q is empty.

(3) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

g(x) → i(x)

The TRS R 2 is

f(x, x) → f(i(x), g(g(x)))
f(x, i(x)) → f(x, x)

The signature Sigma is {f}

(4) Obligation:

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

f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)

The set Q consists of the following terms:

f(x0, x0)
g(x0)
f(x0, i(x0))

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) Obligation:

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

F(x, x) → F(i(x), g(g(x)))
F(x, x) → G(g(x))
F(x, x) → G(x)
F(x, i(x)) → F(x, x)

The TRS R consists of the following rules:

f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)

The set Q consists of the following terms:

f(x0, x0)
g(x0)
f(x0, i(x0))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(8) Obligation:

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

F(x, i(x)) → F(x, x)
F(x, x) → F(i(x), g(g(x)))

The TRS R consists of the following rules:

f(x, x) → f(i(x), g(g(x)))
g(x) → i(x)
f(x, i(x)) → f(x, x)

The set Q consists of the following terms:

f(x0, x0)
g(x0)
f(x0, i(x0))

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

(9) UsableRulesProof (EQUIVALENT transformation)

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.

(10) Obligation:

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

F(x, i(x)) → F(x, x)
F(x, x) → F(i(x), g(g(x)))

The TRS R consists of the following rules:

g(x) → i(x)

The set Q consists of the following terms:

f(x0, x0)
g(x0)
f(x0, i(x0))

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

(11) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

f(x0, x0)
f(x0, i(x0))

(12) Obligation:

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

F(x, i(x)) → F(x, x)
F(x, x) → F(i(x), g(g(x)))

The TRS R consists of the following rules:

g(x) → i(x)

The set Q consists of the following terms:

g(x0)

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

(13) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(x, x) → F(i(x), g(g(x))) at position [1] we obtained the following new rules [LPAR04]:

F(x, x) → F(i(x), i(g(x)))

(14) Obligation:

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

F(x, i(x)) → F(x, x)
F(x, x) → F(i(x), i(g(x)))

The TRS R consists of the following rules:

g(x) → i(x)

The set Q consists of the following terms:

g(x0)

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

(15) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(x, x) → F(i(x), i(g(x))) at position [1,0] we obtained the following new rules [LPAR04]:

F(x, x) → F(i(x), i(i(x)))

(16) Obligation:

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

F(x, i(x)) → F(x, x)
F(x, x) → F(i(x), i(i(x)))

The TRS R consists of the following rules:

g(x) → i(x)

The set Q consists of the following terms:

g(x0)

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

(17) UsableRulesProof (EQUIVALENT transformation)

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:

F(x, i(x)) → F(x, x)
F(x, x) → F(i(x), i(i(x)))

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

g(x0)

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

(19) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

g(x0)

(20) Obligation:

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

F(x, i(x)) → F(x, x)
F(x, x) → F(i(x), i(i(x)))

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

(21) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, i(x)) → F(x, x) we obtained the following new rules [LPAR04]:

F(i(z0), i(i(z0))) → F(i(z0), i(z0))

(22) Obligation:

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

F(x, x) → F(i(x), i(i(x)))
F(i(z0), i(i(z0))) → F(i(z0), i(z0))

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

(23) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, x) → F(i(x), i(i(x))) we obtained the following new rules [LPAR04]:

F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0))))

(24) Obligation:

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

F(i(z0), i(i(z0))) → F(i(z0), i(z0))
F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0))))

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

(25) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(i(z0), i(i(z0))) → F(i(z0), i(z0)) we obtained the following new rules [LPAR04]:

F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))

(26) Obligation:

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

F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0))))
F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))

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

(27) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(i(z0), i(z0)) → F(i(i(z0)), i(i(i(z0)))) we obtained the following new rules [LPAR04]:

F(i(i(z0)), i(i(z0))) → F(i(i(i(z0))), i(i(i(i(z0)))))

(28) Obligation:

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

F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))
F(i(i(z0)), i(i(z0))) → F(i(i(i(z0))), i(i(i(i(z0)))))

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

(29) 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(i(i(z0')), i(i(z0'))) evaluates to t =F(i(i(i(z0'))), i(i(i(z0'))))

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

• Matcher: [z0' / i(z0')]
• Semiunifier: [ ]

---

Rewriting sequence

F(i(i(z0')), i(i(z0'))) → F(i(i(i(z0'))), i(i(i(i(z0')))))
with rule F(i(i(z0'')), i(i(z0''))) → F(i(i(i(z0''))), i(i(i(i(z0''))))) at position [] and matcher [z0'' / z0']

F(i(i(i(z0'))), i(i(i(i(z0'))))) → F(i(i(i(z0'))), i(i(i(z0'))))
with rule F(i(i(z0)), i(i(i(z0)))) → F(i(i(z0)), i(i(z0)))

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.

(30) NO