/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) UsableRulesProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QReductionProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(10) YES


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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, l1)
   ifappend(l1, l2, nil) -> l2
   ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2))

Q is empty.

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(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   hd(cons(x, l)) -> x
   tl(cons(x, l)) -> l
   ifappend(l1, l2, nil) -> l2
   ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2))
   append(l1, l2) -> ifappend(l1, l2, l1)

The TRS R 2 is 
   is_empty(nil) -> true
   is_empty(cons(x, l)) -> false

The signature Sigma is {is_empty_1, true, false}
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(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, l1)
   ifappend(l1, l2, nil) -> l2
   ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, 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, nil)
   ifappend(x0, x1, cons(x2, x3))


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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APPEND(l1, l2) -> IFAPPEND(l1, l2, l1)
   IFAPPEND(l1, l2, cons(x, l)) -> APPEND(l, l2)

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, l1)
   ifappend(l1, l2, nil) -> l2
   ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, 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, nil)
   ifappend(x0, x1, cons(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
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(5) 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.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APPEND(l1, l2) -> IFAPPEND(l1, l2, l1)
   IFAPPEND(l1, l2, cons(x, l)) -> APPEND(l, l2)

R is empty.
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, nil)
   ifappend(x0, x1, cons(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
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(7) 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].

   is_empty(nil)
   is_empty(cons(x0, x1))
   hd(cons(x0, x1))
   tl(cons(x0, x1))
   append(x0, x1)
   ifappend(x0, x1, nil)
   ifappend(x0, x1, cons(x2, x3))


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

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

   APPEND(l1, l2) -> IFAPPEND(l1, l2, l1)
   IFAPPEND(l1, l2, cons(x, l)) -> APPEND(l, l2)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) 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:
*IFAPPEND(l1, l2, cons(x, l)) -> APPEND(l, l2)
The graph contains the following edges 3 > 1, 2 >= 2


*APPEND(l1, l2) -> IFAPPEND(l1, l2, l1)
The graph contains the following edges 1 >= 1, 2 >= 2, 1 >= 3


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(10)
YES