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


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


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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPOrderProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) PisEmptyProof [EQUIVALENT, 0 ms]
        (29) YES


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

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

   msort(nil) -> nil
   msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
   min(x, nil) -> x
   min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
   del(x, nil) -> nil
   del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

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:

   msort(nil) -> nil
   msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
   min(x, nil) -> x
   min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
   del(x, nil) -> nil
   del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

   msort(nil)
   msort(.(x0, x1))
   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))


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

(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:

   MSORT(.(x, y)) -> MIN(x, y)
   MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))
   MSORT(.(x, y)) -> DEL(min(x, y), .(x, y))
   MIN(x, .(y, z)) -> MIN(x, z)
   MIN(x, .(y, z)) -> MIN(y, z)
   DEL(x, .(y, z)) -> DEL(x, z)

The TRS R consists of the following rules:

   msort(nil) -> nil
   msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
   min(x, nil) -> x
   min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
   del(x, nil) -> nil
   del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

   msort(nil)
   msort(.(x0, x1))
   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DEL(x, .(y, z)) -> DEL(x, z)

The TRS R consists of the following rules:

   msort(nil) -> nil
   msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
   min(x, nil) -> x
   min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
   del(x, nil) -> nil
   del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

   msort(nil)
   msort(.(x0, x1))
   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))

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

(8) 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.
----------------------------------------

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

   DEL(x, .(y, z)) -> DEL(x, z)

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

   msort(nil)
   msort(.(x0, x1))
   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))

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

(10) 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].

   msort(nil)
   msort(.(x0, x1))
   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))


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

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

   DEL(x, .(y, z)) -> DEL(x, z)

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

(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:
*DEL(x, .(y, z)) -> DEL(x, z)
The graph contains the following edges 1 >= 1, 2 > 2


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

(13)
YES

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

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

   MIN(x, .(y, z)) -> MIN(y, z)
   MIN(x, .(y, z)) -> MIN(x, z)

The TRS R consists of the following rules:

   msort(nil) -> nil
   msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
   min(x, nil) -> x
   min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
   del(x, nil) -> nil
   del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

   msort(nil)
   msort(.(x0, x1))
   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))

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

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

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

   MIN(x, .(y, z)) -> MIN(y, z)
   MIN(x, .(y, z)) -> MIN(x, z)

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

   msort(nil)
   msort(.(x0, x1))
   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))

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

(17) 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].

   msort(nil)
   msort(.(x0, x1))
   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))


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

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

   MIN(x, .(y, z)) -> MIN(y, z)
   MIN(x, .(y, z)) -> MIN(x, z)

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

(19) 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:
*MIN(x, .(y, z)) -> MIN(y, z)
The graph contains the following edges 2 > 1, 2 > 2


*MIN(x, .(y, z)) -> MIN(x, z)
The graph contains the following edges 1 >= 1, 2 > 2


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

(20)
YES

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

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))

The TRS R consists of the following rules:

   msort(nil) -> nil
   msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
   min(x, nil) -> x
   min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
   del(x, nil) -> nil
   del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

The set Q consists of the following terms:

   msort(nil)
   msort(.(x0, x1))
   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))

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

(22) 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.
----------------------------------------

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

   MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))

The TRS R consists of the following rules:

   min(x, nil) -> x
   min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
   del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))
   del(x, nil) -> nil

The set Q consists of the following terms:

   msort(nil)
   msort(.(x0, x1))
   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))

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

(24) 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].

   msort(nil)
   msort(.(x0, x1))


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

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

   MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))

The TRS R consists of the following rules:

   min(x, nil) -> x
   min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
   del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))
   del(x, nil) -> nil

The set Q consists of the following terms:

   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))

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

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


The following pairs can be oriented strictly and are deleted.

   MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
MSORT(x1)  =  x1

.(x1, x2)  =  .

del(x1, x2)  =  del

if(x1, x2, x3)  =  if


Knuth-Bendix order [KBO] with precedence:del > if

and weight map:

   .=2
   del=1
   if=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))


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

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

   min(x, nil) -> x
   min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
   del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))
   del(x, nil) -> nil

The set Q consists of the following terms:

   min(x0, nil)
   min(x0, .(x1, x2))
   del(x0, nil)
   del(x0, .(x1, x2))

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

(28) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(29)
YES