/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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


MAYBE
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Quasi decreasingness of the given CTRS could not be shown:

(0) CTRS
(1) CTRSToQTRSProof [SOUND, 0 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) TransformationProof [EQUIVALENT, 0 ms]
(14) QDP
(15) UsableRulesProof [EQUIVALENT, 0 ms]
(16) QDP
(17) QReductionProof [EQUIVALENT, 0 ms]
(18) QDP
(19) TransformationProof [EQUIVALENT, 0 ms]
(20) QDP
(21) UsableRulesProof [EQUIVALENT, 0 ms]
(22) QDP
(23) QReductionProof [EQUIVALENT, 0 ms]
(24) QDP
(25) TransformationProof [EQUIVALENT, 0 ms]
(26) QDP
(27) UsableRulesProof [EQUIVALENT, 0 ms]
(28) QDP
(29) QReductionProof [EQUIVALENT, 0 ms]
(30) QDP
(31) TransformationProof [EQUIVALENT, 0 ms]
(32) QDP
(33) UsableRulesProof [EQUIVALENT, 0 ms]
(34) QDP
(35) QReductionProof [EQUIVALENT, 0 ms]
(36) QDP
(37) TransformationProof [EQUIVALENT, 0 ms]
(38) QDP
(39) UsableRulesProof [EQUIVALENT, 0 ms]
(40) QDP
(41) QReductionProof [EQUIVALENT, 0 ms]
(42) QDP
(43) TransformationProof [EQUIVALENT, 0 ms]
(44) QDP
(45) UsableRulesProof [EQUIVALENT, 0 ms]
(46) QDP
(47) QReductionProof [EQUIVALENT, 0 ms]
(48) QDP
(49) TransformationProof [EQUIVALENT, 0 ms]
(50) QDP
(51) TransformationProof [EQUIVALENT, 0 ms]
(52) QDP
(53) NonTerminationLoopProof [COMPLETE, 0 ms]
(54) NO


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

(0)
Obligation:
Conditional term rewrite system:
The TRS R is empty.
The conditional TRS C consists of the following conditional rules:

   a -> a <= b -> x, c -> x
   b -> d <= d -> x, e -> x
   c -> d <= d -> x, e -> x


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

(1) CTRSToQTRSProof (SOUND)
The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC].
----------------------------------------

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

   a -> U1(b)
   U1(x) -> U2(c)
   U2(x) -> a
   b -> U3(d)
   U3(x) -> U4(e)
   U4(x) -> d
   c -> U5(d)
   U5(x) -> U6(e)
   U6(x) -> d

Q is empty.

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

(3) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   b -> U3(d)
   U3(x) -> U4(e)
   U4(x) -> d
   c -> U5(d)
   U5(x) -> U6(e)
   U6(x) -> d

The TRS R 2 is 
   a -> U1(b)
   U1(x) -> U2(c)
   U2(x) -> a

The signature Sigma is {a, U1_1, U2_1}
----------------------------------------

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

   a -> U1(b)
   U1(x) -> U2(c)
   U2(x) -> a
   b -> U3(d)
   U3(x) -> U4(e)
   U4(x) -> d
   c -> U5(d)
   U5(x) -> U6(e)
   U6(x) -> d

The set Q consists of the following terms:

   a
   U1(x0)
   U2(x0)
   b
   U3(x0)
   U4(x0)
   c
   U5(x0)
   U6(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:

   A -> U1^1(b)
   A -> B
   U1^1(x) -> U2^1(c)
   U1^1(x) -> C
   U2^1(x) -> A
   B -> U3^1(d)
   U3^1(x) -> U4^1(e)
   C -> U5^1(d)
   U5^1(x) -> U6^1(e)

The TRS R consists of the following rules:

   a -> U1(b)
   U1(x) -> U2(c)
   U2(x) -> a
   b -> U3(d)
   U3(x) -> U4(e)
   U4(x) -> d
   c -> U5(d)
   U5(x) -> U6(e)
   U6(x) -> d

The set Q consists of the following terms:

   a
   U1(x0)
   U2(x0)
   b
   U3(x0)
   U4(x0)
   c
   U5(x0)
   U6(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 6 less nodes.
----------------------------------------

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

   U1^1(x) -> U2^1(c)
   U2^1(x) -> A
   A -> U1^1(b)

The TRS R consists of the following rules:

   a -> U1(b)
   U1(x) -> U2(c)
   U2(x) -> a
   b -> U3(d)
   U3(x) -> U4(e)
   U4(x) -> d
   c -> U5(d)
   U5(x) -> U6(e)
   U6(x) -> d

The set Q consists of the following terms:

   a
   U1(x0)
   U2(x0)
   b
   U3(x0)
   U4(x0)
   c
   U5(x0)
   U6(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:

   U1^1(x) -> U2^1(c)
   U2^1(x) -> A
   A -> U1^1(b)

The TRS R consists of the following rules:

   b -> U3(d)
   U3(x) -> U4(e)
   U4(x) -> d
   c -> U5(d)
   U5(x) -> U6(e)
   U6(x) -> d

The set Q consists of the following terms:

   a
   U1(x0)
   U2(x0)
   b
   U3(x0)
   U4(x0)
   c
   U5(x0)
   U6(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].

   a
   U1(x0)
   U2(x0)


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

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

   U1^1(x) -> U2^1(c)
   U2^1(x) -> A
   A -> U1^1(b)

The TRS R consists of the following rules:

   b -> U3(d)
   U3(x) -> U4(e)
   U4(x) -> d
   c -> U5(d)
   U5(x) -> U6(e)
   U6(x) -> d

The set Q consists of the following terms:

   b
   U3(x0)
   U4(x0)
   c
   U5(x0)
   U6(x0)

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

(13) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule U1^1(x) -> U2^1(c) at position [0] we obtained the following new rules [LPAR04]:

   (U1^1(x) -> U2^1(U5(d)),U1^1(x) -> U2^1(U5(d)))


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

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

   U2^1(x) -> A
   A -> U1^1(b)
   U1^1(x) -> U2^1(U5(d))

The TRS R consists of the following rules:

   b -> U3(d)
   U3(x) -> U4(e)
   U4(x) -> d
   c -> U5(d)
   U5(x) -> U6(e)
   U6(x) -> d

The set Q consists of the following terms:

   b
   U3(x0)
   U4(x0)
   c
   U5(x0)
   U6(x0)

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

(15) 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.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   A -> U1^1(b)
   U1^1(x) -> U2^1(U5(d))

The TRS R consists of the following rules:

   U5(x) -> U6(e)
   U6(x) -> d
   b -> U3(d)
   U3(x) -> U4(e)
   U4(x) -> d

The set Q consists of the following terms:

   b
   U3(x0)
   U4(x0)
   c
   U5(x0)
   U6(x0)

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

(17) 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].

   c


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

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   A -> U1^1(b)
   U1^1(x) -> U2^1(U5(d))

The TRS R consists of the following rules:

   U5(x) -> U6(e)
   U6(x) -> d
   b -> U3(d)
   U3(x) -> U4(e)
   U4(x) -> d

The set Q consists of the following terms:

   b
   U3(x0)
   U4(x0)
   U5(x0)
   U6(x0)

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

(19) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule A -> U1^1(b) at position [0] we obtained the following new rules [LPAR04]:

   (A -> U1^1(U3(d)),A -> U1^1(U3(d)))


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

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   U1^1(x) -> U2^1(U5(d))
   A -> U1^1(U3(d))

The TRS R consists of the following rules:

   U5(x) -> U6(e)
   U6(x) -> d
   b -> U3(d)
   U3(x) -> U4(e)
   U4(x) -> d

The set Q consists of the following terms:

   b
   U3(x0)
   U4(x0)
   U5(x0)
   U6(x0)

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

(21) 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.
----------------------------------------

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   U1^1(x) -> U2^1(U5(d))
   A -> U1^1(U3(d))

The TRS R consists of the following rules:

   U3(x) -> U4(e)
   U4(x) -> d
   U5(x) -> U6(e)
   U6(x) -> d

The set Q consists of the following terms:

   b
   U3(x0)
   U4(x0)
   U5(x0)
   U6(x0)

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

(23) 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].

   b


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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   U1^1(x) -> U2^1(U5(d))
   A -> U1^1(U3(d))

The TRS R consists of the following rules:

   U3(x) -> U4(e)
   U4(x) -> d
   U5(x) -> U6(e)
   U6(x) -> d

The set Q consists of the following terms:

   U3(x0)
   U4(x0)
   U5(x0)
   U6(x0)

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

(25) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule U1^1(x) -> U2^1(U5(d)) at position [0] we obtained the following new rules [LPAR04]:

   (U1^1(x) -> U2^1(U6(e)),U1^1(x) -> U2^1(U6(e)))


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

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   A -> U1^1(U3(d))
   U1^1(x) -> U2^1(U6(e))

The TRS R consists of the following rules:

   U3(x) -> U4(e)
   U4(x) -> d
   U5(x) -> U6(e)
   U6(x) -> d

The set Q consists of the following terms:

   U3(x0)
   U4(x0)
   U5(x0)
   U6(x0)

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

(27) 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.
----------------------------------------

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   A -> U1^1(U3(d))
   U1^1(x) -> U2^1(U6(e))

The TRS R consists of the following rules:

   U6(x) -> d
   U3(x) -> U4(e)
   U4(x) -> d

The set Q consists of the following terms:

   U3(x0)
   U4(x0)
   U5(x0)
   U6(x0)

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

(29) 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].

   U5(x0)


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

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   A -> U1^1(U3(d))
   U1^1(x) -> U2^1(U6(e))

The TRS R consists of the following rules:

   U6(x) -> d
   U3(x) -> U4(e)
   U4(x) -> d

The set Q consists of the following terms:

   U3(x0)
   U4(x0)
   U6(x0)

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

(31) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule A -> U1^1(U3(d)) at position [0] we obtained the following new rules [LPAR04]:

   (A -> U1^1(U4(e)),A -> U1^1(U4(e)))


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

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   U1^1(x) -> U2^1(U6(e))
   A -> U1^1(U4(e))

The TRS R consists of the following rules:

   U6(x) -> d
   U3(x) -> U4(e)
   U4(x) -> d

The set Q consists of the following terms:

   U3(x0)
   U4(x0)
   U6(x0)

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

(33) 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.
----------------------------------------

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   U1^1(x) -> U2^1(U6(e))
   A -> U1^1(U4(e))

The TRS R consists of the following rules:

   U4(x) -> d
   U6(x) -> d

The set Q consists of the following terms:

   U3(x0)
   U4(x0)
   U6(x0)

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

(35) 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].

   U3(x0)


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

(36)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   U1^1(x) -> U2^1(U6(e))
   A -> U1^1(U4(e))

The TRS R consists of the following rules:

   U4(x) -> d
   U6(x) -> d

The set Q consists of the following terms:

   U4(x0)
   U6(x0)

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

(37) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule U1^1(x) -> U2^1(U6(e)) at position [0] we obtained the following new rules [LPAR04]:

   (U1^1(x) -> U2^1(d),U1^1(x) -> U2^1(d))


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

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   A -> U1^1(U4(e))
   U1^1(x) -> U2^1(d)

The TRS R consists of the following rules:

   U4(x) -> d
   U6(x) -> d

The set Q consists of the following terms:

   U4(x0)
   U6(x0)

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

(39) 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.
----------------------------------------

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   A -> U1^1(U4(e))
   U1^1(x) -> U2^1(d)

The TRS R consists of the following rules:

   U4(x) -> d

The set Q consists of the following terms:

   U4(x0)
   U6(x0)

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

(41) 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].

   U6(x0)


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

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   A -> U1^1(U4(e))
   U1^1(x) -> U2^1(d)

The TRS R consists of the following rules:

   U4(x) -> d

The set Q consists of the following terms:

   U4(x0)

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

(43) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule A -> U1^1(U4(e)) at position [0] we obtained the following new rules [LPAR04]:

   (A -> U1^1(d),A -> U1^1(d))


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

(44)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   U1^1(x) -> U2^1(d)
   A -> U1^1(d)

The TRS R consists of the following rules:

   U4(x) -> d

The set Q consists of the following terms:

   U4(x0)

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

(45) 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.
----------------------------------------

(46)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   U1^1(x) -> U2^1(d)
   A -> U1^1(d)

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

   U4(x0)

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

(47) 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].

   U4(x0)


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

(48)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U2^1(x) -> A
   U1^1(x) -> U2^1(d)
   A -> U1^1(d)

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

(49) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule U2^1(x) -> A we obtained the following new rules [LPAR04]:

   (U2^1(d) -> A,U2^1(d) -> A)


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

(50)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U1^1(x) -> U2^1(d)
   A -> U1^1(d)
   U2^1(d) -> A

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

(51) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule U1^1(x) -> U2^1(d) we obtained the following new rules [LPAR04]:

   (U1^1(d) -> U2^1(d),U1^1(d) -> U2^1(d))


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

(52)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A -> U1^1(d)
   U2^1(d) -> A
   U1^1(d) -> U2^1(d)

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

(53) 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 = U1^1(d) evaluates to  t =U1^1(d)

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

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

U1^1(d) -> U2^1(d)
with rule U1^1(d) -> U2^1(d) at position [] and matcher [ ]

U2^1(d) -> A
with rule U2^1(d) -> A at position [] and matcher [ ]

A -> U1^1(d)
with rule A -> U1^1(d)

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.




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

(54)
NO