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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) UsableRulesProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QReductionProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) TransformationProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) DependencyGraphProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) TransformationProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) DependencyGraphProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) QDPOrderProof [EQUIVALENT, 0 ms]
        (33) QDP
        (34) DependencyGraphProof [EQUIVALENT, 0 ms]
        (35) TRUE


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

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

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
   if_quot(true, x, y) -> s(quot(minus(x, y), y))
   if_quot(false, x, y) -> 0

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)


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

(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:

   MINUS(s(x), s(y)) -> MINUS(x, y)
   LE(s(x), s(y)) -> LE(x, y)
   QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))
   QUOT(x, s(y)) -> LE(s(y), x)
   IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
   IF_QUOT(true, x, y) -> MINUS(x, y)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
   if_quot(true, x, y) -> s(quot(minus(x, y), y))
   if_quot(false, x, y) -> 0

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)

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 2 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   LE(s(x), s(y)) -> LE(x, y)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
   if_quot(true, x, y) -> s(quot(minus(x, y), y))
   if_quot(false, x, y) -> 0

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)

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:

   LE(s(x), s(y)) -> LE(x, y)

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

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)

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].

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)


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

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

   LE(s(x), s(y)) -> LE(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:
*LE(s(x), s(y)) -> LE(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:

   MINUS(s(x), s(y)) -> MINUS(x, y)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
   if_quot(true, x, y) -> s(quot(minus(x, y), y))
   if_quot(false, x, y) -> 0

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)

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:

   MINUS(s(x), s(y)) -> MINUS(x, y)

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

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)

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].

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)


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

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

   MINUS(s(x), s(y)) -> 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:
*MINUS(s(x), s(y)) -> 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:

   IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
   QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
   if_quot(true, x, y) -> s(quot(minus(x, y), y))
   if_quot(false, x, y) -> 0

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)

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

(20) 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.
----------------------------------------

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

   IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
   QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   le(0, y) -> true
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)

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

(22) 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].

   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)


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

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

   IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
   QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   le(0, y) -> true
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

(24) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y)) at position [0] we obtained the following new rules [LPAR04]:

   (QUOT(0, s(x0)) -> IF_QUOT(false, 0, s(x0)),QUOT(0, s(x0)) -> IF_QUOT(false, 0, s(x0)))
   (QUOT(s(x1), s(x0)) -> IF_QUOT(le(x0, x1), s(x1), s(x0)),QUOT(s(x1), s(x0)) -> IF_QUOT(le(x0, x1), s(x1), s(x0)))


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

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

   IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
   QUOT(0, s(x0)) -> IF_QUOT(false, 0, s(x0))
   QUOT(s(x1), s(x0)) -> IF_QUOT(le(x0, x1), s(x1), s(x0))

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   le(0, y) -> true
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

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

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

   QUOT(s(x1), s(x0)) -> IF_QUOT(le(x0, x1), s(x1), s(x0))
   IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   le(0, y) -> true
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

(28) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule IF_QUOT(true, x, y) -> QUOT(minus(x, y), y) at position [0] we obtained the following new rules [LPAR04]:

   (IF_QUOT(true, x0, 0) -> QUOT(x0, 0),IF_QUOT(true, x0, 0) -> QUOT(x0, 0))
   (IF_QUOT(true, s(x0), s(x1)) -> QUOT(minus(x0, x1), s(x1)),IF_QUOT(true, s(x0), s(x1)) -> QUOT(minus(x0, x1), s(x1)))


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

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   QUOT(s(x1), s(x0)) -> IF_QUOT(le(x0, x1), s(x1), s(x0))
   IF_QUOT(true, x0, 0) -> QUOT(x0, 0)
   IF_QUOT(true, s(x0), s(x1)) -> QUOT(minus(x0, x1), s(x1))

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   le(0, y) -> true
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

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

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF_QUOT(true, s(x0), s(x1)) -> QUOT(minus(x0, x1), s(x1))
   QUOT(s(x1), s(x0)) -> IF_QUOT(le(x0, x1), s(x1), s(x0))

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   le(0, y) -> true
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

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


The following pairs can be oriented strictly and are deleted.

   IF_QUOT(true, s(x0), s(x1)) -> QUOT(minus(x0, x1), s(x1))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
IF_QUOT(x1, x2, x3)  =  x2

s(x1)  =  s(x1)

QUOT(x1, x2)  =  x1

minus(x1, x2)  =  x1


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s_1=1
   dummyConstant=1

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

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)


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

(33)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   QUOT(s(x1), s(x0)) -> IF_QUOT(le(x0, x1), s(x1), s(x0))

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   le(0, y) -> true
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

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

(35)
TRUE