/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 63 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 20 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 0 ms]
(6) QTRS
(7) QTRSRRRProof [EQUIVALENT, 10 ms]
(8) QTRS
(9) QTRSRRRProof [EQUIVALENT, 11 ms]
(10) QTRS
(11) DependencyPairsProof [EQUIVALENT, 0 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 5 ms]
(14) QDP
(15) MRRProof [EQUIVALENT, 0 ms]
(16) QDP
(17) DependencyGraphProof [EQUIVALENT, 0 ms]
(18) QDP
(19) QDPOrderProof [EQUIVALENT, 39 ms]
(20) QDP
(21) DependencyGraphProof [EQUIVALENT, 0 ms]
(22) QDP
(23) UsableRulesProof [EQUIVALENT, 0 ms]
(24) QDP
(25) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(26) YES


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

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

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   a__head(cons(X, XS)) -> mark(X)
   a__tail(cons(X, XS)) -> mark(XS)
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(head(X)) -> a__head(mark(X))
   mark(tail(X)) -> a__tail(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(nil) -> nil
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__nats -> nats
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)
   a__head(X) -> head(X)
   a__tail(X) -> tail(X)

Q is empty.

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

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

   POL(0) = 0
   POL(a__head(x_1)) = 1 + x_1
   POL(a__incr(x_1)) = 2*x_1
   POL(a__nats) = 0
   POL(a__odds) = 0
   POL(a__pairs) = 0
   POL(a__tail(x_1)) = 1 + 2*x_1
   POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(head(x_1)) = 1 + x_1
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = 2*x_1
   POL(nats) = 0
   POL(nil) = 0
   POL(odds) = 0
   POL(pairs) = 0
   POL(s(x_1)) = x_1
   POL(tail(x_1)) = 1 + 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   a__head(cons(X, XS)) -> mark(X)
   a__tail(cons(X, XS)) -> mark(XS)
   mark(head(X)) -> a__head(mark(X))
   mark(tail(X)) -> a__tail(mark(X))




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

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

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(nil) -> nil
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__nats -> nats
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)
   a__head(X) -> head(X)
   a__tail(X) -> tail(X)

Q is empty.

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

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

   POL(0) = 0
   POL(a__head(x_1)) = 1 + x_1
   POL(a__incr(x_1)) = 2*x_1
   POL(a__nats) = 0
   POL(a__odds) = 0
   POL(a__pairs) = 0
   POL(a__tail(x_1)) = 2 + x_1
   POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(head(x_1)) = x_1
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = 2*x_1
   POL(nats) = 0
   POL(nil) = 0
   POL(odds) = 0
   POL(pairs) = 0
   POL(s(x_1)) = x_1
   POL(tail(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:

   a__head(X) -> head(X)




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

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

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(nil) -> nil
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__nats -> nats
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)
   a__tail(X) -> tail(X)

Q is empty.

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

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

   POL(0) = 0
   POL(a__incr(x_1)) = 2*x_1
   POL(a__nats) = 0
   POL(a__odds) = 0
   POL(a__pairs) = 0
   POL(a__tail(x_1)) = 1 + x_1
   POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = 2*x_1
   POL(nats) = 0
   POL(nil) = 0
   POL(odds) = 0
   POL(pairs) = 0
   POL(s(x_1)) = x_1
   POL(tail(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   a__tail(X) -> tail(X)




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

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

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(nil) -> nil
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__nats -> nats
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)

Q is empty.

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

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

   POL(0) = 0
   POL(a__incr(x_1)) = 2*x_1
   POL(a__nats) = 0
   POL(a__odds) = 0
   POL(a__pairs) = 0
   POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = 2*x_1
   POL(nats) = 0
   POL(nil) = 1
   POL(odds) = 0
   POL(pairs) = 0
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   mark(nil) -> nil




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

(8)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__nats -> nats
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)

Q is empty.

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

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

   POL(0) = 0
   POL(a__incr(x_1)) = 2*x_1
   POL(a__nats) = 2
   POL(a__odds) = 0
   POL(a__pairs) = 0
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = 2*x_1
   POL(nats) = 1
   POL(odds) = 0
   POL(pairs) = 0
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   a__nats -> nats




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

(10)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)

Q is empty.

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

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

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

   A__ODDS -> A__INCR(a__pairs)
   A__ODDS -> A__PAIRS
   A__INCR(cons(X, XS)) -> MARK(X)
   MARK(nats) -> A__NATS
   MARK(pairs) -> A__PAIRS
   MARK(odds) -> A__ODDS
   MARK(incr(X)) -> A__INCR(mark(X))
   MARK(incr(X)) -> MARK(X)
   MARK(s(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> MARK(X1)

The TRS R consists of the following rules:

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)

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

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

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

   A__INCR(cons(X, XS)) -> MARK(X)
   MARK(odds) -> A__ODDS
   A__ODDS -> A__INCR(a__pairs)
   MARK(incr(X)) -> A__INCR(mark(X))
   MARK(incr(X)) -> MARK(X)
   MARK(s(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> MARK(X1)

The TRS R consists of the following rules:

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)

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

(15) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   MARK(odds) -> A__ODDS


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(A__INCR(x_1)) = x_1
   POL(A__ODDS) = 2
   POL(MARK(x_1)) = 2*x_1
   POL(a__incr(x_1)) = x_1
   POL(a__nats) = 0
   POL(a__odds) = 2
   POL(a__pairs) = 2
   POL(cons(x_1, x_2)) = 2*x_1 + x_2
   POL(incr(x_1)) = x_1
   POL(mark(x_1)) = x_1
   POL(nats) = 0
   POL(odds) = 2
   POL(pairs) = 2
   POL(s(x_1)) = x_1


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

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

   A__INCR(cons(X, XS)) -> MARK(X)
   A__ODDS -> A__INCR(a__pairs)
   MARK(incr(X)) -> A__INCR(mark(X))
   MARK(incr(X)) -> MARK(X)
   MARK(s(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> MARK(X1)

The TRS R consists of the following rules:

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)

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

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

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

   MARK(incr(X)) -> A__INCR(mark(X))
   A__INCR(cons(X, XS)) -> MARK(X)
   MARK(incr(X)) -> MARK(X)
   MARK(s(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> MARK(X1)

The TRS R consists of the following rules:

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)

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

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


The following pairs can be oriented strictly and are deleted.

   MARK(incr(X)) -> A__INCR(mark(X))
   MARK(cons(X1, X2)) -> MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( A__INCR_1(x_1) ) = max{0, 2x_1 - 1}
POL( mark_1(x_1) ) = x_1
POL( nats ) = 1
POL( a__nats ) = 1
POL( pairs ) = 1
POL( a__pairs ) = 1
POL( odds ) = 2
POL( a__odds ) = 2
POL( incr_1(x_1) ) = 2x_1
POL( a__incr_1(x_1) ) = 2x_1
POL( 0 ) = 0
POL( s_1(x_1) ) = x_1
POL( cons_2(x_1, x_2) ) = 2x_1 + 1
POL( MARK_1(x_1) ) = x_1 + 1

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

   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__odds -> odds
   a__odds -> a__incr(a__pairs)
   a__pairs -> cons(0, incr(odds))
   a__pairs -> pairs
   a__incr(X) -> incr(X)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   a__nats -> cons(0, incr(nats))


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

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

   A__INCR(cons(X, XS)) -> MARK(X)
   MARK(incr(X)) -> MARK(X)
   MARK(s(X)) -> MARK(X)

The TRS R consists of the following rules:

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)

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

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

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

   MARK(s(X)) -> MARK(X)
   MARK(incr(X)) -> MARK(X)

The TRS R consists of the following rules:

   a__nats -> cons(0, incr(nats))
   a__pairs -> cons(0, incr(odds))
   a__odds -> a__incr(a__pairs)
   a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS))
   mark(nats) -> a__nats
   mark(pairs) -> a__pairs
   mark(odds) -> a__odds
   mark(incr(X)) -> a__incr(mark(X))
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   a__pairs -> pairs
   a__odds -> odds
   a__incr(X) -> incr(X)

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

(23) 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.
----------------------------------------

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(s(X)) -> MARK(X)
   MARK(incr(X)) -> MARK(X)

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

(25) 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:
*MARK(s(X)) -> MARK(X)
The graph contains the following edges 1 > 1


*MARK(incr(X)) -> MARK(X)
The graph contains the following edges 1 > 1


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

(26)
YES