NO
proof of Strategy_removed_CSR_05_ExIntrod_GM01.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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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, 69 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 0 ms]
(4) QTRS
(5) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 0 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) AND
    (11) QDP
        (12) UsableRulesProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QReductionProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) NonTerminationLoopProof [COMPLETE, 0 ms]
        (17) NO
    (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QReductionProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) UsableRulesProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QReductionProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (31) YES


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

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

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

Q is empty.

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

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

   POL(0) = 0
   POL(adx(x_1)) = 2*x_1
   POL(cons(x_1, x_2)) = x_1 + 2*x_2
   POL(head(x_1)) = 2 + x_1
   POL(incr(x_1)) = x_1
   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)
   head(cons(X, L)) -> X
   tail(cons(X, L)) -> 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), incr(L))
   adx(nil) -> nil
   adx(cons(X, L)) -> incr(cons(X, adx(L)))
   zeros -> cons(0, zeros)

Q is empty.

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

(3) 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 + x_2
   POL(incr(x_1)) = x_1
   POL(nil) = 1
   POL(s(x_1)) = x_1
   POL(zeros) = 2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   adx(nil) -> nil




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

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

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

Q is empty.

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

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

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

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

The set Q consists of the following terms:

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


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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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

(10)
Complex Obligation (AND)

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

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

   ZEROS -> ZEROS

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(12) 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.
----------------------------------------

(13)
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:

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

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

(14) 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].

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


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

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

(16) 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: [ ]

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

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




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

(17)
NO

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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(19) 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.
----------------------------------------

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

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

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

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

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

(21) 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].

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


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

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

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

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

(23) 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, L)) -> INCR(L)
The graph contains the following edges 1 > 1


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

(24)
YES

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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(26) 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.
----------------------------------------

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

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

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

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

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

(28) 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].

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


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

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

(30) 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, L)) -> ADX(L)
The graph contains the following edges 1 > 1


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(31)
YES