NO
proof of Strategy_removed_CSR_05_ExIntrod_GM04.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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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, 64 ms]
(2) QTRS
(3) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) UsableRulesProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QReductionProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QReductionProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (22) YES
    (23) QDP
        (24) UsableRulesProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QReductionProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) NonTerminationLoopProof [COMPLETE, 0 ms]
        (29) NO


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

   nats -> adx(zeros)
   zeros -> cons(0, zeros)
   incr(cons(X, Y)) -> cons(s(X), incr(Y))
   adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
   hd(cons(X, Y)) -> X
   tl(cons(X, Y)) -> Y

Q is empty.

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

   POL(0) = 0
   POL(adx(x_1)) = 2*x_1
   POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(hd(x_1)) = 2 + x_1
   POL(incr(x_1)) = x_1
   POL(nats) = 2
   POL(s(x_1)) = x_1
   POL(tl(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)
   hd(cons(X, Y)) -> X
   tl(cons(X, Y)) -> Y




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

   zeros -> cons(0, zeros)
   incr(cons(X, Y)) -> cons(s(X), incr(Y))
   adx(cons(X, Y)) -> incr(cons(X, adx(Y)))

Q is empty.

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(3) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

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

   zeros -> cons(0, zeros)
   incr(cons(X, Y)) -> cons(s(X), incr(Y))
   adx(cons(X, Y)) -> incr(cons(X, adx(Y)))

The set Q consists of the following terms:

   zeros
   incr(cons(x0, x1))
   adx(cons(x0, x1))


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(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:

   ZEROS -> ZEROS
   INCR(cons(X, Y)) -> INCR(Y)
   ADX(cons(X, Y)) -> INCR(cons(X, adx(Y)))
   ADX(cons(X, Y)) -> ADX(Y)

The TRS R consists of the following rules:

   zeros -> cons(0, zeros)
   incr(cons(X, Y)) -> cons(s(X), incr(Y))
   adx(cons(X, Y)) -> incr(cons(X, adx(Y)))

The set Q consists of the following terms:

   zeros
   incr(cons(x0, x1))
   adx(cons(x0, x1))

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

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 1 less node.
----------------------------------------

(8)
Complex Obligation (AND)

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

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INCR(cons(X, Y)) -> INCR(Y)

The TRS R consists of the following rules:

   zeros -> cons(0, zeros)
   incr(cons(X, Y)) -> cons(s(X), incr(Y))
   adx(cons(X, Y)) -> incr(cons(X, adx(Y)))

The set Q consists of the following terms:

   zeros
   incr(cons(x0, x1))
   adx(cons(x0, x1))

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

(10) 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.
----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INCR(cons(X, Y)) -> INCR(Y)

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

   zeros
   incr(cons(x0, x1))
   adx(cons(x0, x1))

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

(12) 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].

   zeros
   incr(cons(x0, x1))
   adx(cons(x0, x1))


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

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INCR(cons(X, Y)) -> INCR(Y)

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

(14) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*INCR(cons(X, Y)) -> INCR(Y)
The graph contains the following edges 1 > 1


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(15)
YES

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

   ADX(cons(X, Y)) -> ADX(Y)

The TRS R consists of the following rules:

   zeros -> cons(0, zeros)
   incr(cons(X, Y)) -> cons(s(X), incr(Y))
   adx(cons(X, Y)) -> incr(cons(X, adx(Y)))

The set Q consists of the following terms:

   zeros
   incr(cons(x0, x1))
   adx(cons(x0, x1))

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

(17) 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.
----------------------------------------

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

   ADX(cons(X, Y)) -> ADX(Y)

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

   zeros
   incr(cons(x0, x1))
   adx(cons(x0, x1))

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

(19) 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].

   zeros
   incr(cons(x0, x1))
   adx(cons(x0, x1))


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

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

   ADX(cons(X, Y)) -> ADX(Y)

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

(21) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*ADX(cons(X, Y)) -> ADX(Y)
The graph contains the following edges 1 > 1


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(22)
YES

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(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ZEROS -> ZEROS

The TRS R consists of the following rules:

   zeros -> cons(0, zeros)
   incr(cons(X, Y)) -> cons(s(X), incr(Y))
   adx(cons(X, Y)) -> incr(cons(X, adx(Y)))

The set Q consists of the following terms:

   zeros
   incr(cons(x0, x1))
   adx(cons(x0, x1))

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

(24) 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.
----------------------------------------

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ZEROS -> ZEROS

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

   zeros
   incr(cons(x0, x1))
   adx(cons(x0, x1))

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

(26) 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].

   zeros
   incr(cons(x0, x1))
   adx(cons(x0, x1))


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

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ZEROS -> ZEROS

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

(28) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = ZEROS evaluates to  t =ZEROS

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

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from ZEROS to ZEROS.




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(29)
NO