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


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YES
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 proven:

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 19 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QReductionProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QDPOrderProof [EQUIVALENT, 29 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 0 ms]
(14) TRUE


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

   is_empty(nil) -> true
   is_empty(cons(x, l)) -> false
   hd(cons(x, l)) -> x
   tl(cons(x, l)) -> l
   append(l1, l2) -> ifappend(l1, l2, is_empty(l1))
   ifappend(l1, l2, true) -> l2
   ifappend(l1, l2, false) -> cons(hd(l1), append(tl(l1), l2))

Q is empty.

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

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

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

   is_empty(nil) -> true
   is_empty(cons(x, l)) -> false
   hd(cons(x, l)) -> x
   tl(cons(x, l)) -> l
   append(l1, l2) -> ifappend(l1, l2, is_empty(l1))
   ifappend(l1, l2, true) -> l2
   ifappend(l1, l2, false) -> cons(hd(l1), append(tl(l1), l2))

The set Q consists of the following terms:

   is_empty(nil)
   is_empty(cons(x0, x1))
   hd(cons(x0, x1))
   tl(cons(x0, x1))
   append(x0, x1)
   ifappend(x0, x1, true)
   ifappend(x0, x1, false)


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

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

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APPEND(l1, l2) -> IFAPPEND(l1, l2, is_empty(l1))
   APPEND(l1, l2) -> IS_EMPTY(l1)
   IFAPPEND(l1, l2, false) -> HD(l1)
   IFAPPEND(l1, l2, false) -> APPEND(tl(l1), l2)
   IFAPPEND(l1, l2, false) -> TL(l1)

The TRS R consists of the following rules:

   is_empty(nil) -> true
   is_empty(cons(x, l)) -> false
   hd(cons(x, l)) -> x
   tl(cons(x, l)) -> l
   append(l1, l2) -> ifappend(l1, l2, is_empty(l1))
   ifappend(l1, l2, true) -> l2
   ifappend(l1, l2, false) -> cons(hd(l1), append(tl(l1), l2))

The set Q consists of the following terms:

   is_empty(nil)
   is_empty(cons(x0, x1))
   hd(cons(x0, x1))
   tl(cons(x0, x1))
   append(x0, x1)
   ifappend(x0, x1, true)
   ifappend(x0, x1, false)

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFAPPEND(l1, l2, false) -> APPEND(tl(l1), l2)
   APPEND(l1, l2) -> IFAPPEND(l1, l2, is_empty(l1))

The TRS R consists of the following rules:

   is_empty(nil) -> true
   is_empty(cons(x, l)) -> false
   hd(cons(x, l)) -> x
   tl(cons(x, l)) -> l
   append(l1, l2) -> ifappend(l1, l2, is_empty(l1))
   ifappend(l1, l2, true) -> l2
   ifappend(l1, l2, false) -> cons(hd(l1), append(tl(l1), l2))

The set Q consists of the following terms:

   is_empty(nil)
   is_empty(cons(x0, x1))
   hd(cons(x0, x1))
   tl(cons(x0, x1))
   append(x0, x1)
   ifappend(x0, x1, true)
   ifappend(x0, x1, false)

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

(7) 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.
----------------------------------------

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

   IFAPPEND(l1, l2, false) -> APPEND(tl(l1), l2)
   APPEND(l1, l2) -> IFAPPEND(l1, l2, is_empty(l1))

The TRS R consists of the following rules:

   is_empty(nil) -> true
   is_empty(cons(x, l)) -> false
   tl(cons(x, l)) -> l

The set Q consists of the following terms:

   is_empty(nil)
   is_empty(cons(x0, x1))
   hd(cons(x0, x1))
   tl(cons(x0, x1))
   append(x0, x1)
   ifappend(x0, x1, true)
   ifappend(x0, x1, false)

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

(9) 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].

   hd(cons(x0, x1))
   append(x0, x1)
   ifappend(x0, x1, true)
   ifappend(x0, x1, false)


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

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

   IFAPPEND(l1, l2, false) -> APPEND(tl(l1), l2)
   APPEND(l1, l2) -> IFAPPEND(l1, l2, is_empty(l1))

The TRS R consists of the following rules:

   is_empty(nil) -> true
   is_empty(cons(x, l)) -> false
   tl(cons(x, l)) -> l

The set Q consists of the following terms:

   is_empty(nil)
   is_empty(cons(x0, x1))
   tl(cons(x0, x1))

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

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


The following pairs can be oriented strictly and are deleted.

   APPEND(l1, l2) -> IFAPPEND(l1, l2, is_empty(l1))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(APPEND(x_1, x_2)) = [1/2] + [2]x_1
   POL(IFAPPEND(x_1, x_2, x_3)) = x_1 + [1/4]x_3
   POL(cons(x_1, x_2)) = [2] + [2]x_2
   POL(false) = [4]
   POL(is_empty(x_1)) = [1/4] + [4]x_1
   POL(nil) = [2]
   POL(tl(x_1)) = [1/4] + [1/2]x_1
   POL(true) = [2]
The value of delta used in the strict ordering is 7/16.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   tl(cons(x, l)) -> l
   is_empty(nil) -> true
   is_empty(cons(x, l)) -> false


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

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

   IFAPPEND(l1, l2, false) -> APPEND(tl(l1), l2)

The TRS R consists of the following rules:

   is_empty(nil) -> true
   is_empty(cons(x, l)) -> false
   tl(cons(x, l)) -> l

The set Q consists of the following terms:

   is_empty(nil)
   is_empty(cons(x0, x1))
   tl(cons(x0, x1))

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

(13) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(14)
TRUE