YES
proof of AProVE_04_Liveness6.2.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 63 ms]
(2) QTRS
(3) QTRSRRRProof [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) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) RFCMatchBoundsDPProof [EQUIVALENT, 0 ms]
        (18) YES


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

   top(sent(x)) -> top(check(rest(x)))
   rest(nil) -> sent(nil)
   rest(cons(x, y)) -> sent(y)
   check(sent(x)) -> sent(check(x))
   check(rest(x)) -> rest(check(x))
   check(cons(x, y)) -> cons(check(x), y)
   check(cons(x, y)) -> cons(x, check(y))
   check(cons(x, y)) -> cons(x, y)

Q is empty.

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

   POL(check(x_1)) = x_1
   POL(cons(x_1, x_2)) = 1 + x_1 + x_2
   POL(nil) = 1
   POL(rest(x_1)) = 2 + 2*x_1
   POL(sent(x_1)) = 2 + 2*x_1
   POL(top(x_1)) = 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   rest(cons(x, y)) -> sent(y)




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

   top(sent(x)) -> top(check(rest(x)))
   rest(nil) -> sent(nil)
   check(sent(x)) -> sent(check(x))
   check(rest(x)) -> rest(check(x))
   check(cons(x, y)) -> cons(check(x), y)
   check(cons(x, y)) -> cons(x, check(y))
   check(cons(x, y)) -> cons(x, y)

Q is empty.

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

   POL(check(x_1)) = 2*x_1
   POL(cons(x_1, x_2)) = 1 + x_1 + x_2
   POL(nil) = 0
   POL(rest(x_1)) = x_1
   POL(sent(x_1)) = 2*x_1
   POL(top(x_1)) = 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   check(cons(x, y)) -> cons(check(x), y)
   check(cons(x, y)) -> cons(x, check(y))
   check(cons(x, y)) -> cons(x, y)




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

   top(sent(x)) -> top(check(rest(x)))
   rest(nil) -> sent(nil)
   check(sent(x)) -> sent(check(x))
   check(rest(x)) -> rest(check(x))

Q is empty.

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(5) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TOP(sent(x)) -> TOP(check(rest(x)))
   TOP(sent(x)) -> CHECK(rest(x))
   TOP(sent(x)) -> REST(x)
   CHECK(sent(x)) -> CHECK(x)
   CHECK(rest(x)) -> REST(check(x))
   CHECK(rest(x)) -> CHECK(x)

The TRS R consists of the following rules:

   top(sent(x)) -> top(check(rest(x)))
   rest(nil) -> sent(nil)
   check(sent(x)) -> sent(check(x))
   check(rest(x)) -> rest(check(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
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(8)
Complex Obligation (AND)

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

   CHECK(rest(x)) -> CHECK(x)
   CHECK(sent(x)) -> CHECK(x)

The TRS R consists of the following rules:

   top(sent(x)) -> top(check(rest(x)))
   rest(nil) -> sent(nil)
   check(sent(x)) -> sent(check(x))
   check(rest(x)) -> rest(check(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(10) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
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(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CHECK(rest(x)) -> CHECK(x)
   CHECK(sent(x)) -> CHECK(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(12) 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:
*CHECK(rest(x)) -> CHECK(x)
The graph contains the following edges 1 > 1


*CHECK(sent(x)) -> CHECK(x)
The graph contains the following edges 1 > 1


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(13)
YES

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

   TOP(sent(x)) -> TOP(check(rest(x)))

The TRS R consists of the following rules:

   top(sent(x)) -> top(check(rest(x)))
   rest(nil) -> sent(nil)
   check(sent(x)) -> sent(check(x))
   check(rest(x)) -> rest(check(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(15) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   TOP(sent(x)) -> TOP(check(rest(x)))

The TRS R consists of the following rules:

   rest(nil) -> sent(nil)
   check(sent(x)) -> sent(check(x))
   check(rest(x)) -> rest(check(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(17) RFCMatchBoundsDPProof (EQUIVALENT)
Finiteness of the DP problem can be shown by a matchbound of 2. 
As the DP problem is minimal we only have to initialize the certificate graph by the rules of P:

   TOP(sent(x)) -> TOP(check(rest(x)))

To find matches we regarded all rules of R and P:

   rest(nil) -> sent(nil)
   check(sent(x)) -> sent(check(x))
   check(rest(x)) -> rest(check(x))
   TOP(sent(x)) -> TOP(check(rest(x)))

The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
55, 56, 57, 58, 59, 60, 61, 62, 63, 64

Node 55 is start node and node 56 is final node.

Those nodes are connected through the following edges:

* 55 to 57 labelled TOP_1(0)* 55 to 62 labelled TOP_1(1)* 56 to 56 labelled #_1(0)* 57 to 58 labelled check_1(0)* 57 to 59 labelled rest_1(1)* 57 to 61 labelled sent_1(1)* 58 to 56 labelled rest_1(0)* 58 to 60 labelled sent_1(1)* 59 to 56 labelled check_1(1)* 59 to 59 labelled sent_1(1), rest_1(1)* 60 to 56 labelled nil(1)* 61 to 60 labelled check_1(1)* 62 to 63 labelled check_1(1)* 62 to 64 labelled rest_1(2)* 63 to 61 labelled rest_1(1)* 64 to 61 labelled check_1(2)


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(18)
YES