YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QReductionProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QReductionProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) QDPOrderProof [EQUIVALENT, 88 ms]
        (21) QDP
        (22) DependencyGraphProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (25) YES


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

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

   a__minus(0, Y) -> 0
   a__minus(s(X), s(Y)) -> a__minus(X, Y)
   a__geq(X, 0) -> true
   a__geq(0, s(Y)) -> false
   a__geq(s(X), s(Y)) -> a__geq(X, Y)
   a__div(0, s(Y)) -> 0
   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   a__if(true, X, Y) -> mark(X)
   a__if(false, X, Y) -> mark(Y)
   mark(minus(X1, X2)) -> a__minus(X1, X2)
   mark(geq(X1, X2)) -> a__geq(X1, X2)
   mark(div(X1, X2)) -> a__div(mark(X1), X2)
   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(true) -> true
   mark(false) -> false
   a__minus(X1, X2) -> minus(X1, X2)
   a__geq(X1, X2) -> geq(X1, X2)
   a__div(X1, X2) -> div(X1, X2)
   a__if(X1, X2, X3) -> if(X1, X2, X3)

The set Q consists of the following terms:

   mark(minus(x0, x1))
   mark(geq(x0, x1))
   mark(div(x0, x1))
   mark(if(x0, x1, x2))
   mark(0)
   mark(s(x0))
   mark(true)
   mark(false)
   a__minus(x0, x1)
   a__geq(x0, x1)
   a__div(x0, x1)
   a__if(x0, x1, x2)


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

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

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__MINUS(s(X), s(Y)) -> A__MINUS(X, Y)
   A__GEQ(s(X), s(Y)) -> A__GEQ(X, Y)
   A__DIV(s(X), s(Y)) -> A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   A__DIV(s(X), s(Y)) -> A__GEQ(X, Y)
   A__IF(true, X, Y) -> MARK(X)
   A__IF(false, X, Y) -> MARK(Y)
   MARK(minus(X1, X2)) -> A__MINUS(X1, X2)
   MARK(geq(X1, X2)) -> A__GEQ(X1, X2)
   MARK(div(X1, X2)) -> A__DIV(mark(X1), X2)
   MARK(div(X1, X2)) -> MARK(X1)
   MARK(if(X1, X2, X3)) -> A__IF(mark(X1), X2, X3)
   MARK(if(X1, X2, X3)) -> MARK(X1)
   MARK(s(X)) -> MARK(X)

The TRS R consists of the following rules:

   a__minus(0, Y) -> 0
   a__minus(s(X), s(Y)) -> a__minus(X, Y)
   a__geq(X, 0) -> true
   a__geq(0, s(Y)) -> false
   a__geq(s(X), s(Y)) -> a__geq(X, Y)
   a__div(0, s(Y)) -> 0
   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   a__if(true, X, Y) -> mark(X)
   a__if(false, X, Y) -> mark(Y)
   mark(minus(X1, X2)) -> a__minus(X1, X2)
   mark(geq(X1, X2)) -> a__geq(X1, X2)
   mark(div(X1, X2)) -> a__div(mark(X1), X2)
   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(true) -> true
   mark(false) -> false
   a__minus(X1, X2) -> minus(X1, X2)
   a__geq(X1, X2) -> geq(X1, X2)
   a__div(X1, X2) -> div(X1, X2)
   a__if(X1, X2, X3) -> if(X1, X2, X3)

The set Q consists of the following terms:

   mark(minus(x0, x1))
   mark(geq(x0, x1))
   mark(div(x0, x1))
   mark(if(x0, x1, x2))
   mark(0)
   mark(s(x0))
   mark(true)
   mark(false)
   a__minus(x0, x1)
   a__geq(x0, x1)
   a__div(x0, x1)
   a__if(x0, x1, x2)

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

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

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__GEQ(s(X), s(Y)) -> A__GEQ(X, Y)

The TRS R consists of the following rules:

   a__minus(0, Y) -> 0
   a__minus(s(X), s(Y)) -> a__minus(X, Y)
   a__geq(X, 0) -> true
   a__geq(0, s(Y)) -> false
   a__geq(s(X), s(Y)) -> a__geq(X, Y)
   a__div(0, s(Y)) -> 0
   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   a__if(true, X, Y) -> mark(X)
   a__if(false, X, Y) -> mark(Y)
   mark(minus(X1, X2)) -> a__minus(X1, X2)
   mark(geq(X1, X2)) -> a__geq(X1, X2)
   mark(div(X1, X2)) -> a__div(mark(X1), X2)
   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(true) -> true
   mark(false) -> false
   a__minus(X1, X2) -> minus(X1, X2)
   a__geq(X1, X2) -> geq(X1, X2)
   a__div(X1, X2) -> div(X1, X2)
   a__if(X1, X2, X3) -> if(X1, X2, X3)

The set Q consists of the following terms:

   mark(minus(x0, x1))
   mark(geq(x0, x1))
   mark(div(x0, x1))
   mark(if(x0, x1, x2))
   mark(0)
   mark(s(x0))
   mark(true)
   mark(false)
   a__minus(x0, x1)
   a__geq(x0, x1)
   a__div(x0, x1)
   a__if(x0, x1, x2)

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

(6) 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.
----------------------------------------

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

   A__GEQ(s(X), s(Y)) -> A__GEQ(X, Y)

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

   mark(minus(x0, x1))
   mark(geq(x0, x1))
   mark(div(x0, x1))
   mark(if(x0, x1, x2))
   mark(0)
   mark(s(x0))
   mark(true)
   mark(false)
   a__minus(x0, x1)
   a__geq(x0, x1)
   a__div(x0, x1)
   a__if(x0, x1, x2)

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

(8) 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].

   mark(minus(x0, x1))
   mark(geq(x0, x1))
   mark(div(x0, x1))
   mark(if(x0, x1, x2))
   mark(0)
   mark(s(x0))
   mark(true)
   mark(false)
   a__minus(x0, x1)
   a__geq(x0, x1)
   a__div(x0, x1)
   a__if(x0, x1, x2)


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

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

   A__GEQ(s(X), s(Y)) -> A__GEQ(X, Y)

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

(10) 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:
*A__GEQ(s(X), s(Y)) -> A__GEQ(X, Y)
The graph contains the following edges 1 > 1, 2 > 2


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

(11)
YES

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

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

   A__MINUS(s(X), s(Y)) -> A__MINUS(X, Y)

The TRS R consists of the following rules:

   a__minus(0, Y) -> 0
   a__minus(s(X), s(Y)) -> a__minus(X, Y)
   a__geq(X, 0) -> true
   a__geq(0, s(Y)) -> false
   a__geq(s(X), s(Y)) -> a__geq(X, Y)
   a__div(0, s(Y)) -> 0
   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   a__if(true, X, Y) -> mark(X)
   a__if(false, X, Y) -> mark(Y)
   mark(minus(X1, X2)) -> a__minus(X1, X2)
   mark(geq(X1, X2)) -> a__geq(X1, X2)
   mark(div(X1, X2)) -> a__div(mark(X1), X2)
   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(true) -> true
   mark(false) -> false
   a__minus(X1, X2) -> minus(X1, X2)
   a__geq(X1, X2) -> geq(X1, X2)
   a__div(X1, X2) -> div(X1, X2)
   a__if(X1, X2, X3) -> if(X1, X2, X3)

The set Q consists of the following terms:

   mark(minus(x0, x1))
   mark(geq(x0, x1))
   mark(div(x0, x1))
   mark(if(x0, x1, x2))
   mark(0)
   mark(s(x0))
   mark(true)
   mark(false)
   a__minus(x0, x1)
   a__geq(x0, x1)
   a__div(x0, x1)
   a__if(x0, x1, x2)

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

(13) 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.
----------------------------------------

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

   A__MINUS(s(X), s(Y)) -> A__MINUS(X, Y)

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

   mark(minus(x0, x1))
   mark(geq(x0, x1))
   mark(div(x0, x1))
   mark(if(x0, x1, x2))
   mark(0)
   mark(s(x0))
   mark(true)
   mark(false)
   a__minus(x0, x1)
   a__geq(x0, x1)
   a__div(x0, x1)
   a__if(x0, x1, x2)

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

(15) 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].

   mark(minus(x0, x1))
   mark(geq(x0, x1))
   mark(div(x0, x1))
   mark(if(x0, x1, x2))
   mark(0)
   mark(s(x0))
   mark(true)
   mark(false)
   a__minus(x0, x1)
   a__geq(x0, x1)
   a__div(x0, x1)
   a__if(x0, x1, x2)


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

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

   A__MINUS(s(X), s(Y)) -> A__MINUS(X, Y)

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

(17) 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:
*A__MINUS(s(X), s(Y)) -> A__MINUS(X, Y)
The graph contains the following edges 1 > 1, 2 > 2


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

(18)
YES

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

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(div(X1, X2)) -> A__DIV(mark(X1), X2)
   A__DIV(s(X), s(Y)) -> A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   A__IF(true, X, Y) -> MARK(X)
   MARK(div(X1, X2)) -> MARK(X1)
   MARK(if(X1, X2, X3)) -> A__IF(mark(X1), X2, X3)
   A__IF(false, X, Y) -> MARK(Y)
   MARK(if(X1, X2, X3)) -> MARK(X1)
   MARK(s(X)) -> MARK(X)

The TRS R consists of the following rules:

   a__minus(0, Y) -> 0
   a__minus(s(X), s(Y)) -> a__minus(X, Y)
   a__geq(X, 0) -> true
   a__geq(0, s(Y)) -> false
   a__geq(s(X), s(Y)) -> a__geq(X, Y)
   a__div(0, s(Y)) -> 0
   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   a__if(true, X, Y) -> mark(X)
   a__if(false, X, Y) -> mark(Y)
   mark(minus(X1, X2)) -> a__minus(X1, X2)
   mark(geq(X1, X2)) -> a__geq(X1, X2)
   mark(div(X1, X2)) -> a__div(mark(X1), X2)
   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(true) -> true
   mark(false) -> false
   a__minus(X1, X2) -> minus(X1, X2)
   a__geq(X1, X2) -> geq(X1, X2)
   a__div(X1, X2) -> div(X1, X2)
   a__if(X1, X2, X3) -> if(X1, X2, X3)

The set Q consists of the following terms:

   mark(minus(x0, x1))
   mark(geq(x0, x1))
   mark(div(x0, x1))
   mark(if(x0, x1, x2))
   mark(0)
   mark(s(x0))
   mark(true)
   mark(false)
   a__minus(x0, x1)
   a__geq(x0, x1)
   a__div(x0, x1)
   a__if(x0, x1, x2)

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

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


The following pairs can be oriented strictly and are deleted.

   A__DIV(s(X), s(Y)) -> A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   MARK(s(X)) -> MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( A__DIV_2(x_1, x_2) ) = 2x_1 + 2x_2 + 2
POL( A__IF_3(x_1, ..., x_3) ) = x_2 + x_3 + 2
POL( mark_1(x_1) ) = x_1
POL( minus_2(x_1, x_2) ) = 0
POL( a__minus_2(x_1, x_2) ) = 0
POL( geq_2(x_1, x_2) ) = 0
POL( a__geq_2(x_1, x_2) ) = 0
POL( div_2(x_1, x_2) ) = 2x_1 + 2x_2
POL( a__div_2(x_1, x_2) ) = 2x_1 + 2x_2
POL( s_1(x_1) ) = x_1 + 1
POL( a__if_3(x_1, ..., x_3) ) = x_1 + x_2 + 2x_3
POL( 0 ) = 0
POL( true ) = 0
POL( if_3(x_1, ..., x_3) ) = x_1 + x_2 + 2x_3
POL( false ) = 0
POL( MARK_1(x_1) ) = x_1 + 2

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

   mark(minus(X1, X2)) -> a__minus(X1, X2)
   mark(geq(X1, X2)) -> a__geq(X1, X2)
   mark(div(X1, X2)) -> a__div(mark(X1), X2)
   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   a__if(true, X, Y) -> mark(X)
   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
   a__if(false, X, Y) -> mark(Y)
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(true) -> true
   mark(false) -> false
   a__geq(X, 0) -> true
   a__geq(0, s(Y)) -> false
   a__geq(s(X), s(Y)) -> a__geq(X, Y)
   a__geq(X1, X2) -> geq(X1, X2)
   a__div(0, s(Y)) -> 0
   a__div(X1, X2) -> div(X1, X2)
   a__if(X1, X2, X3) -> if(X1, X2, X3)
   a__minus(0, Y) -> 0
   a__minus(s(X), s(Y)) -> a__minus(X, Y)
   a__minus(X1, X2) -> minus(X1, X2)


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

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

   MARK(div(X1, X2)) -> A__DIV(mark(X1), X2)
   A__IF(true, X, Y) -> MARK(X)
   MARK(div(X1, X2)) -> MARK(X1)
   MARK(if(X1, X2, X3)) -> A__IF(mark(X1), X2, X3)
   A__IF(false, X, Y) -> MARK(Y)
   MARK(if(X1, X2, X3)) -> MARK(X1)

The TRS R consists of the following rules:

   a__minus(0, Y) -> 0
   a__minus(s(X), s(Y)) -> a__minus(X, Y)
   a__geq(X, 0) -> true
   a__geq(0, s(Y)) -> false
   a__geq(s(X), s(Y)) -> a__geq(X, Y)
   a__div(0, s(Y)) -> 0
   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   a__if(true, X, Y) -> mark(X)
   a__if(false, X, Y) -> mark(Y)
   mark(minus(X1, X2)) -> a__minus(X1, X2)
   mark(geq(X1, X2)) -> a__geq(X1, X2)
   mark(div(X1, X2)) -> a__div(mark(X1), X2)
   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(true) -> true
   mark(false) -> false
   a__minus(X1, X2) -> minus(X1, X2)
   a__geq(X1, X2) -> geq(X1, X2)
   a__div(X1, X2) -> div(X1, X2)
   a__if(X1, X2, X3) -> if(X1, X2, X3)

The set Q consists of the following terms:

   mark(minus(x0, x1))
   mark(geq(x0, x1))
   mark(div(x0, x1))
   mark(if(x0, x1, x2))
   mark(0)
   mark(s(x0))
   mark(true)
   mark(false)
   a__minus(x0, x1)
   a__geq(x0, x1)
   a__div(x0, x1)
   a__if(x0, x1, x2)

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

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

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

   MARK(div(X1, X2)) -> MARK(X1)
   MARK(if(X1, X2, X3)) -> A__IF(mark(X1), X2, X3)
   A__IF(true, X, Y) -> MARK(X)
   MARK(if(X1, X2, X3)) -> MARK(X1)
   A__IF(false, X, Y) -> MARK(Y)

The TRS R consists of the following rules:

   a__minus(0, Y) -> 0
   a__minus(s(X), s(Y)) -> a__minus(X, Y)
   a__geq(X, 0) -> true
   a__geq(0, s(Y)) -> false
   a__geq(s(X), s(Y)) -> a__geq(X, Y)
   a__div(0, s(Y)) -> 0
   a__div(s(X), s(Y)) -> a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   a__if(true, X, Y) -> mark(X)
   a__if(false, X, Y) -> mark(Y)
   mark(minus(X1, X2)) -> a__minus(X1, X2)
   mark(geq(X1, X2)) -> a__geq(X1, X2)
   mark(div(X1, X2)) -> a__div(mark(X1), X2)
   mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
   mark(0) -> 0
   mark(s(X)) -> s(mark(X))
   mark(true) -> true
   mark(false) -> false
   a__minus(X1, X2) -> minus(X1, X2)
   a__geq(X1, X2) -> geq(X1, X2)
   a__div(X1, X2) -> div(X1, X2)
   a__if(X1, X2, X3) -> if(X1, X2, X3)

The set Q consists of the following terms:

   mark(minus(x0, x1))
   mark(geq(x0, x1))
   mark(div(x0, x1))
   mark(if(x0, x1, x2))
   mark(0)
   mark(s(x0))
   mark(true)
   mark(false)
   a__minus(x0, x1)
   a__geq(x0, x1)
   a__div(x0, x1)
   a__if(x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
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(24) 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(if(X1, X2, X3)) -> A__IF(mark(X1), X2, X3)
The graph contains the following edges 1 > 2, 1 > 3


*MARK(div(X1, X2)) -> MARK(X1)
The graph contains the following edges 1 > 1


*MARK(if(X1, X2, X3)) -> MARK(X1)
The graph contains the following edges 1 > 1


*A__IF(true, X, Y) -> MARK(X)
The graph contains the following edges 2 >= 1


*A__IF(false, X, Y) -> MARK(Y)
The graph contains the following edges 3 >= 1


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(25)
YES