NO
proof of Strategy_removed_mixed_05_ex4.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 21 ms]
(2) QTRS
(3) AAECC Innermost [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) UsableRulesProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QReductionProof [EQUIVALENT, 0 ms]
(12) QDP
(13) MNOCProof [EQUIVALENT, 0 ms]
(14) QDP
(15) NonTerminationLoopProof [COMPLETE, 0 ms]
(16) NO


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

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

   f(h(x)) -> f(i(x))
   f(i(x)) -> a
   i(x) -> h(x)

Q is empty.

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

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(a) = 0
   POL(f(x_1)) = 1 + x_1
   POL(h(x_1)) = x_1
   POL(i(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   f(i(x)) -> a




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

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(h(x)) -> f(i(x))
   i(x) -> h(x)

Q is empty.

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

(3) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   i(x) -> h(x)

The TRS R 2 is 
   f(h(x)) -> f(i(x))

The signature Sigma is {f_1}
----------------------------------------

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(h(x)) -> f(i(x))
   i(x) -> h(x)

The set Q consists of the following terms:

   f(h(x0))
   i(x0)


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

(5) DependencyPairsProof (EQUIVALENT)
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(h(x)) -> F(i(x))
   F(h(x)) -> I(x)

The TRS R consists of the following rules:

   f(h(x)) -> f(i(x))
   i(x) -> h(x)

The set Q consists of the following terms:

   f(h(x0))
   i(x0)

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

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(h(x)) -> F(i(x))

The TRS R consists of the following rules:

   f(h(x)) -> f(i(x))
   i(x) -> h(x)

The set Q consists of the following terms:

   f(h(x0))
   i(x0)

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

(9) UsableRulesProof (EQUIVALENT)
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(h(x)) -> F(i(x))

The TRS R consists of the following rules:

   i(x) -> h(x)

The set Q consists of the following terms:

   f(h(x0))
   i(x0)

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

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

   f(h(x0))


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

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(h(x)) -> F(i(x))

The TRS R consists of the following rules:

   i(x) -> h(x)

The set Q consists of the following terms:

   i(x0)

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

(13) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
----------------------------------------

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(h(x)) -> F(i(x))

The TRS R consists of the following rules:

   i(x) -> h(x)

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

(15) NonTerminationLoopProof (COMPLETE)
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(x')) evaluates to  t =F(i(x'))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [ ]

--------------------------------------------------------------------------------
Rewriting sequence

F(i(x')) -> F(h(x'))
with rule i(x'') -> h(x'') at position [0] and matcher [x'' / x']

F(h(x')) -> F(i(x'))
with rule F(h(x)) -> F(i(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.




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

(16)
NO