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


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


NO
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 72 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 17 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 0 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 0 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) MRRProof [EQUIVALENT, 15 ms]
(12) QDP
(13) QDPOrderProof [EQUIVALENT, 42 ms]
(14) QDP
(15) TransformationProof [EQUIVALENT, 0 ms]
(16) QDP
(17) TransformationProof [EQUIVALENT, 0 ms]
(18) QDP
(19) TransformationProof [EQUIVALENT, 0 ms]
(20) QDP
(21) DependencyGraphProof [EQUIVALENT, 0 ms]
(22) QDP
(23) TransformationProof [EQUIVALENT, 0 ms]
(24) QDP
(25) DependencyGraphProof [EQUIVALENT, 0 ms]
(26) QDP
(27) QDPOrderProof [EQUIVALENT, 30 ms]
(28) QDP
(29) QDPOrderProof [EQUIVALENT, 398 ms]
(30) QDP
(31) NonTerminationLoopProof [COMPLETE, 33 ms]
(32) NO


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

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

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(nil) -> nil
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   nats -> adx(zeros)
   zeros -> cons(0, n__zeros)
   head(cons(X, L)) -> X
   tail(cons(X, L)) -> activate(L)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

Q is empty.

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

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

   POL(0) = 0
   POL(activate(x_1)) = x_1
   POL(adx(x_1)) = x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(head(x_1)) = 2*x_1
   POL(incr(x_1)) = x_1
   POL(n__adx(x_1)) = x_1
   POL(n__incr(x_1)) = x_1
   POL(n__zeros) = 0
   POL(nats) = 2
   POL(nil) = 0
   POL(s(x_1)) = x_1
   POL(tail(x_1)) = 2 + 2*x_1
   POL(zeros) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   nats -> adx(zeros)
   tail(cons(X, L)) -> activate(L)




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

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

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(nil) -> nil
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   head(cons(X, L)) -> X
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

Q is empty.

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

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

   POL(0) = 0
   POL(activate(x_1)) = x_1
   POL(adx(x_1)) = x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(head(x_1)) = 1 + 2*x_1
   POL(incr(x_1)) = x_1
   POL(n__adx(x_1)) = x_1
   POL(n__incr(x_1)) = x_1
   POL(n__zeros) = 0
   POL(nil) = 0
   POL(s(x_1)) = x_1
   POL(zeros) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   head(cons(X, L)) -> X




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

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

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(nil) -> nil
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

Q is empty.

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

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

   POL(0) = 0
   POL(activate(x_1)) = x_1
   POL(adx(x_1)) = 1 + x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(incr(x_1)) = x_1
   POL(n__adx(x_1)) = 1 + x_1
   POL(n__incr(x_1)) = x_1
   POL(n__zeros) = 0
   POL(nil) = 0
   POL(s(x_1)) = x_1
   POL(zeros) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   adx(nil) -> nil




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

(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

Q is empty.

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

(7) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

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

   INCR(cons(X, L)) -> ACTIVATE(L)
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))
   ADX(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__incr(X)) -> INCR(activate(X))
   ACTIVATE(n__incr(X)) -> ACTIVATE(X)
   ACTIVATE(n__adx(X)) -> ADX(activate(X))
   ACTIVATE(n__adx(X)) -> ACTIVATE(X)
   ACTIVATE(n__zeros) -> ZEROS

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

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

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVATE(n__incr(X)) -> INCR(activate(X))
   INCR(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__incr(X)) -> ACTIVATE(X)
   ACTIVATE(n__adx(X)) -> ADX(activate(X))
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))
   ADX(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__adx(X)) -> ACTIVATE(X)

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

(11) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   ADX(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__adx(X)) -> ACTIVATE(X)


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(ACTIVATE(x_1)) = x_1
   POL(ADX(x_1)) = 1 + 2*x_1
   POL(INCR(x_1)) = x_1
   POL(activate(x_1)) = x_1
   POL(adx(x_1)) = 1 + 2*x_1
   POL(cons(x_1, x_2)) = 2*x_1 + x_2
   POL(incr(x_1)) = x_1
   POL(n__adx(x_1)) = 1 + 2*x_1
   POL(n__incr(x_1)) = x_1
   POL(n__zeros) = 0
   POL(nil) = 0
   POL(s(x_1)) = x_1
   POL(zeros) = 0


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

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

   ACTIVATE(n__incr(X)) -> INCR(activate(X))
   INCR(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__incr(X)) -> ACTIVATE(X)
   ACTIVATE(n__adx(X)) -> ADX(activate(X))
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

(13) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   ACTIVATE(n__incr(X)) -> ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( ADX_1(x_1) ) = 2
POL( INCR_1(x_1) ) = 2x_1 + 2
POL( activate_1(x_1) ) = x_1
POL( n__incr_1(x_1) ) = x_1 + 1
POL( incr_1(x_1) ) = x_1 + 1
POL( n__adx_1(x_1) ) = 1
POL( adx_1(x_1) ) = 1
POL( n__zeros ) = 1
POL( zeros ) = 1
POL( nil ) = 0
POL( cons_2(x_1, x_2) ) = max{0, x_2 - 1}
POL( s_1(x_1) ) = x_1 + 2
POL( 0 ) = 2
POL( ACTIVATE_1(x_1) ) = 2x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X
   adx(X) -> n__adx(X)
   incr(nil) -> nil
   incr(X) -> n__incr(X)
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   zeros -> n__zeros


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

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

   ACTIVATE(n__incr(X)) -> INCR(activate(X))
   INCR(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__adx(X)) -> ADX(activate(X))
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

(15) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule ACTIVATE(n__incr(X)) -> INCR(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

   (ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0))),ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0))))
   (ACTIVATE(n__incr(n__adx(x0))) -> INCR(adx(activate(x0))),ACTIVATE(n__incr(n__adx(x0))) -> INCR(adx(activate(x0))))
   (ACTIVATE(n__incr(n__zeros)) -> INCR(zeros),ACTIVATE(n__incr(n__zeros)) -> INCR(zeros))
   (ACTIVATE(n__incr(x0)) -> INCR(x0),ACTIVATE(n__incr(x0)) -> INCR(x0))


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

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

   INCR(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__adx(X)) -> ADX(activate(X))
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))
   ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0)))
   ACTIVATE(n__incr(n__adx(x0))) -> INCR(adx(activate(x0)))
   ACTIVATE(n__incr(n__zeros)) -> INCR(zeros)
   ACTIVATE(n__incr(x0)) -> INCR(x0)

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

(17) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule ACTIVATE(n__adx(X)) -> ADX(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

   (ACTIVATE(n__adx(n__incr(x0))) -> ADX(incr(activate(x0))),ACTIVATE(n__adx(n__incr(x0))) -> ADX(incr(activate(x0))))
   (ACTIVATE(n__adx(n__adx(x0))) -> ADX(adx(activate(x0))),ACTIVATE(n__adx(n__adx(x0))) -> ADX(adx(activate(x0))))
   (ACTIVATE(n__adx(n__zeros)) -> ADX(zeros),ACTIVATE(n__adx(n__zeros)) -> ADX(zeros))
   (ACTIVATE(n__adx(x0)) -> ADX(x0),ACTIVATE(n__adx(x0)) -> ADX(x0))


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

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

   INCR(cons(X, L)) -> ACTIVATE(L)
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))
   ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0)))
   ACTIVATE(n__incr(n__adx(x0))) -> INCR(adx(activate(x0)))
   ACTIVATE(n__incr(n__zeros)) -> INCR(zeros)
   ACTIVATE(n__incr(x0)) -> INCR(x0)
   ACTIVATE(n__adx(n__incr(x0))) -> ADX(incr(activate(x0)))
   ACTIVATE(n__adx(n__adx(x0))) -> ADX(adx(activate(x0)))
   ACTIVATE(n__adx(n__zeros)) -> ADX(zeros)
   ACTIVATE(n__adx(x0)) -> ADX(x0)

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

(19) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule ACTIVATE(n__incr(n__zeros)) -> INCR(zeros) at position [0] we obtained the following new rules [LPAR04]:

   (ACTIVATE(n__incr(n__zeros)) -> INCR(cons(0, n__zeros)),ACTIVATE(n__incr(n__zeros)) -> INCR(cons(0, n__zeros)))
   (ACTIVATE(n__incr(n__zeros)) -> INCR(n__zeros),ACTIVATE(n__incr(n__zeros)) -> INCR(n__zeros))


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

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

   INCR(cons(X, L)) -> ACTIVATE(L)
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))
   ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0)))
   ACTIVATE(n__incr(n__adx(x0))) -> INCR(adx(activate(x0)))
   ACTIVATE(n__incr(x0)) -> INCR(x0)
   ACTIVATE(n__adx(n__incr(x0))) -> ADX(incr(activate(x0)))
   ACTIVATE(n__adx(n__adx(x0))) -> ADX(adx(activate(x0)))
   ACTIVATE(n__adx(n__zeros)) -> ADX(zeros)
   ACTIVATE(n__adx(x0)) -> ADX(x0)
   ACTIVATE(n__incr(n__zeros)) -> INCR(cons(0, n__zeros))
   ACTIVATE(n__incr(n__zeros)) -> INCR(n__zeros)

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

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

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

   ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0)))
   INCR(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__incr(n__adx(x0))) -> INCR(adx(activate(x0)))
   ACTIVATE(n__incr(x0)) -> INCR(x0)
   ACTIVATE(n__adx(n__incr(x0))) -> ADX(incr(activate(x0)))
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))
   ACTIVATE(n__adx(n__adx(x0))) -> ADX(adx(activate(x0)))
   ACTIVATE(n__adx(n__zeros)) -> ADX(zeros)
   ACTIVATE(n__adx(x0)) -> ADX(x0)
   ACTIVATE(n__incr(n__zeros)) -> INCR(cons(0, n__zeros))

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

(23) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule ACTIVATE(n__adx(n__zeros)) -> ADX(zeros) at position [0] we obtained the following new rules [LPAR04]:

   (ACTIVATE(n__adx(n__zeros)) -> ADX(cons(0, n__zeros)),ACTIVATE(n__adx(n__zeros)) -> ADX(cons(0, n__zeros)))
   (ACTIVATE(n__adx(n__zeros)) -> ADX(n__zeros),ACTIVATE(n__adx(n__zeros)) -> ADX(n__zeros))


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

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

   ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0)))
   INCR(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__incr(n__adx(x0))) -> INCR(adx(activate(x0)))
   ACTIVATE(n__incr(x0)) -> INCR(x0)
   ACTIVATE(n__adx(n__incr(x0))) -> ADX(incr(activate(x0)))
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))
   ACTIVATE(n__adx(n__adx(x0))) -> ADX(adx(activate(x0)))
   ACTIVATE(n__adx(x0)) -> ADX(x0)
   ACTIVATE(n__incr(n__zeros)) -> INCR(cons(0, n__zeros))
   ACTIVATE(n__adx(n__zeros)) -> ADX(cons(0, n__zeros))
   ACTIVATE(n__adx(n__zeros)) -> ADX(n__zeros)

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

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

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

   INCR(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0)))
   ACTIVATE(n__incr(n__adx(x0))) -> INCR(adx(activate(x0)))
   ACTIVATE(n__incr(x0)) -> INCR(x0)
   ACTIVATE(n__adx(n__incr(x0))) -> ADX(incr(activate(x0)))
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))
   ACTIVATE(n__adx(n__adx(x0))) -> ADX(adx(activate(x0)))
   ACTIVATE(n__adx(x0)) -> ADX(x0)
   ACTIVATE(n__incr(n__zeros)) -> INCR(cons(0, n__zeros))
   ACTIVATE(n__adx(n__zeros)) -> ADX(cons(0, n__zeros))

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

(27) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   ACTIVATE(n__incr(n__zeros)) -> INCR(cons(0, n__zeros))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( ADX_1(x_1) ) = max{0, -2}
POL( INCR_1(x_1) ) = x_1
POL( activate_1(x_1) ) = x_1
POL( n__incr_1(x_1) ) = 2x_1
POL( incr_1(x_1) ) = 2x_1
POL( n__adx_1(x_1) ) = max{0, -2}
POL( adx_1(x_1) ) = 0
POL( cons_2(x_1, x_2) ) = x_2
POL( n__zeros ) = 2
POL( zeros ) = 2
POL( nil ) = 1
POL( s_1(x_1) ) = x_1 + 2
POL( 0 ) = 2
POL( ACTIVATE_1(x_1) ) = x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X
   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   incr(X) -> n__incr(X)
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   adx(X) -> n__adx(X)
   zeros -> cons(0, n__zeros)
   zeros -> n__zeros


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

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

   INCR(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0)))
   ACTIVATE(n__incr(n__adx(x0))) -> INCR(adx(activate(x0)))
   ACTIVATE(n__incr(x0)) -> INCR(x0)
   ACTIVATE(n__adx(n__incr(x0))) -> ADX(incr(activate(x0)))
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))
   ACTIVATE(n__adx(n__adx(x0))) -> ADX(adx(activate(x0)))
   ACTIVATE(n__adx(x0)) -> ADX(x0)
   ACTIVATE(n__adx(n__zeros)) -> ADX(cons(0, n__zeros))

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

(29) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   ACTIVATE(n__adx(n__adx(x0))) -> ADX(adx(activate(x0)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(0) = 0
   POL(ACTIVATE(x_1)) = x_1
   POL(ADX(x_1)) = [1/2] + [1/4]x_1
   POL(INCR(x_1)) = [1/4]x_1
   POL(activate(x_1)) = x_1
   POL(adx(x_1)) = [1/2] + x_1
   POL(cons(x_1, x_2)) = [4]x_2
   POL(incr(x_1)) = [1/4]x_1
   POL(n__adx(x_1)) = [1/2] + x_1
   POL(n__incr(x_1)) = [1/4]x_1
   POL(n__zeros) = 0
   POL(nil) = 0
   POL(s(x_1)) = 0
   POL(zeros) = 0
The value of delta used in the strict ordering is 3/8.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X
   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   incr(X) -> n__incr(X)
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   adx(X) -> n__adx(X)
   zeros -> cons(0, n__zeros)
   zeros -> n__zeros


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

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

   INCR(cons(X, L)) -> ACTIVATE(L)
   ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0)))
   ACTIVATE(n__incr(n__adx(x0))) -> INCR(adx(activate(x0)))
   ACTIVATE(n__incr(x0)) -> INCR(x0)
   ACTIVATE(n__adx(n__incr(x0))) -> ADX(incr(activate(x0)))
   ADX(cons(X, L)) -> INCR(cons(X, n__adx(activate(L))))
   ACTIVATE(n__adx(x0)) -> ADX(x0)
   ACTIVATE(n__adx(n__zeros)) -> ADX(cons(0, n__zeros))

The TRS R consists of the following rules:

   incr(nil) -> nil
   incr(cons(X, L)) -> cons(s(X), n__incr(activate(L)))
   adx(cons(X, L)) -> incr(cons(X, n__adx(activate(L))))
   zeros -> cons(0, n__zeros)
   incr(X) -> n__incr(X)
   adx(X) -> n__adx(X)
   zeros -> n__zeros
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__adx(X)) -> adx(activate(X))
   activate(n__zeros) -> zeros
   activate(X) -> X

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

(31) 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 = ACTIVATE(n__adx(activate(n__zeros))) evaluates to  t =ACTIVATE(n__adx(activate(n__zeros)))

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

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

ACTIVATE(n__adx(activate(n__zeros))) -> ACTIVATE(n__adx(n__zeros))
with rule activate(X) -> X at position [0,0] and matcher [X / n__zeros]

ACTIVATE(n__adx(n__zeros)) -> ADX(cons(0, n__zeros))
with rule ACTIVATE(n__adx(n__zeros)) -> ADX(cons(0, n__zeros)) at position [] and matcher [ ]

ADX(cons(0, n__zeros)) -> INCR(cons(0, n__adx(activate(n__zeros))))
with rule ADX(cons(X', L')) -> INCR(cons(X', n__adx(activate(L')))) at position [] and matcher [X' / 0, L' / n__zeros]

INCR(cons(0, n__adx(activate(n__zeros)))) -> ACTIVATE(n__adx(activate(n__zeros)))
with rule INCR(cons(X, L)) -> ACTIVATE(L)

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.




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

(32)
NO