/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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, 2 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) NonInfProof [EQUIVALENT, 147 ms]
        (27) AND
            (28) QDP
                (29) DependencyGraphProof [EQUIVALENT, 0 ms]
                (30) TRUE
            (31) QDP
                (32) DependencyGraphProof [EQUIVALENT, 0 ms]
                (33) TRUE


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

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

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(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:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, 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:

   LE(s(x), s(y)) -> LE(x, y)
   DOUBLE(s(x)) -> DOUBLE(x)
   LOG(s(x)) -> LOOP(s(x), s(0), 0)
   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)
   LOOP(x, s(y), z) -> LE(x, s(y))
   IF(false, x, y, z) -> LOOP(x, double(y), s(z))
   IF(false, x, y, z) -> DOUBLE(y)

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, 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 3 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   DOUBLE(s(x)) -> DOUBLE(x)

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, 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:

   DOUBLE(s(x)) -> DOUBLE(x)

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

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, 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].

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)


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

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

   DOUBLE(s(x)) -> DOUBLE(x)

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:
*DOUBLE(s(x)) -> DOUBLE(x)
The graph contains the following edges 1 > 1


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

(13)
YES

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

(14)
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:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, 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:

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

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

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, 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].

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)


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

(18)
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.
----------------------------------------

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

   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)
   IF(false, x, y, z) -> LOOP(x, double(y), s(z))

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, 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:

   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)
   IF(false, x, y, z) -> LOOP(x, double(y), s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, 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].

   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)


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

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

   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)
   IF(false, x, y, z) -> LOOP(x, double(y), s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(26) NonInfProof (EQUIVALENT)
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:

Note that final constraints are written in bold face.



For Pair LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z) the following chains were created:
*We consider the chain LOOP(x3, s(x4), x5) -> IF(le(x3, s(x4)), x3, s(x4), x5), IF(false, x6, x7, x8) -> LOOP(x6, double(x7), s(x8)) which results in the following constraint:

(1)    (IF(le(x3, s(x4)), x3, s(x4), x5)=IF(false, x6, x7, x8)  ==>  LOOP(x3, s(x4), x5)_>=_IF(le(x3, s(x4)), x3, s(x4), x5))



We simplified constraint (1) using rules (I), (II), (IV), (VII) which results in the following new constraint:

(2)    (s(x4)=x18 & le(x3, x18)=false  ==>  LOOP(x3, s(x4), x5)_>=_IF(le(x3, s(x4)), x3, s(x4), x5))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x3, x18)=false which results in the following new constraints:

(3)    (le(x21, x20)=false & s(x4)=s(x20) & (\/x22,x23:le(x21, x20)=false & s(x22)=x20  ==>  LOOP(x21, s(x22), x23)_>=_IF(le(x21, s(x22)), x21, s(x22), x23))  ==>  LOOP(s(x21), s(x4), x5)_>=_IF(le(s(x21), s(x4)), s(x21), s(x4), x5))

(4)    (false=false & s(x4)=0  ==>  LOOP(s(x24), s(x4), x5)_>=_IF(le(s(x24), s(x4)), s(x24), s(x4), x5))



We simplified constraint (3) using rules (I), (II), (III), (IV) which results in the following new constraint:

(5)    (le(x21, x20)=false  ==>  LOOP(s(x21), s(x20), x5)_>=_IF(le(s(x21), s(x20)), s(x21), s(x20), x5))



We solved constraint (4) using rules (I), (II).We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on le(x21, x20)=false which results in the following new constraints:

(6)    (le(x27, x26)=false & (\/x28:le(x27, x26)=false  ==>  LOOP(s(x27), s(x26), x28)_>=_IF(le(s(x27), s(x26)), s(x27), s(x26), x28))  ==>  LOOP(s(s(x27)), s(s(x26)), x5)_>=_IF(le(s(s(x27)), s(s(x26))), s(s(x27)), s(s(x26)), x5))

(7)    (false=false  ==>  LOOP(s(s(x29)), s(0), x5)_>=_IF(le(s(s(x29)), s(0)), s(s(x29)), s(0), x5))



We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (\/x28:le(x27, x26)=false  ==>  LOOP(s(x27), s(x26), x28)_>=_IF(le(s(x27), s(x26)), s(x27), s(x26), x28)) with sigma = [x28 / x5] which results in the following new constraint:

(8)    (LOOP(s(x27), s(x26), x5)_>=_IF(le(s(x27), s(x26)), s(x27), s(x26), x5)  ==>  LOOP(s(s(x27)), s(s(x26)), x5)_>=_IF(le(s(s(x27)), s(s(x26))), s(s(x27)), s(s(x26)), x5))



We simplified constraint (7) using rules (I), (II) which results in the following new constraint:

(9)    (LOOP(s(s(x29)), s(0), x5)_>=_IF(le(s(s(x29)), s(0)), s(s(x29)), s(0), x5))








For Pair IF(false, x, y, z) -> LOOP(x, double(y), s(z)) the following chains were created:
*We consider the chain IF(false, x9, x10, x11) -> LOOP(x9, double(x10), s(x11)), LOOP(x12, s(x13), x14) -> IF(le(x12, s(x13)), x12, s(x13), x14) which results in the following constraint:

(1)    (LOOP(x9, double(x10), s(x11))=LOOP(x12, s(x13), x14)  ==>  IF(false, x9, x10, x11)_>=_LOOP(x9, double(x10), s(x11)))



We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:

(2)    (double(x10)=s(x13)  ==>  IF(false, x9, x10, x11)_>=_LOOP(x9, double(x10), s(x11)))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on double(x10)=s(x13) which results in the following new constraint:

(3)    (s(s(double(x30)))=s(x13) & (\/x31,x32,x33:double(x30)=s(x31)  ==>  IF(false, x32, x30, x33)_>=_LOOP(x32, double(x30), s(x33)))  ==>  IF(false, x9, s(x30), x11)_>=_LOOP(x9, double(s(x30)), s(x11)))



We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:

(4)    (IF(false, x9, s(x30), x11)_>=_LOOP(x9, double(s(x30)), s(x11)))








To summarize, we get the following constraints P__>=_ for the following pairs.

*LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)

*(LOOP(s(x27), s(x26), x5)_>=_IF(le(s(x27), s(x26)), s(x27), s(x26), x5)  ==>  LOOP(s(s(x27)), s(s(x26)), x5)_>=_IF(le(s(s(x27)), s(s(x26))), s(s(x27)), s(s(x26)), x5))


*(LOOP(s(s(x29)), s(0), x5)_>=_IF(le(s(s(x29)), s(0)), s(s(x29)), s(0), x5))




*IF(false, x, y, z) -> LOOP(x, double(y), s(z))

*(IF(false, x9, s(x30), x11)_>=_LOOP(x9, double(s(x30)), s(x11)))








The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by  "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. 

Using the following integer polynomial ordering the  resulting constraints can be solved 

Polynomial interpretation [NONINF]:

   POL(0) = 0
   POL(IF(x_1, x_2, x_3, x_4)) = -1 - x_1 + x_2 - x_3
   POL(LOOP(x_1, x_2, x_3)) = -1 + x_1 - x_2
   POL(c) = -2
   POL(double(x_1)) = 2*x_1
   POL(false) = 0
   POL(le(x_1, x_2)) = 0
   POL(s(x_1)) = 1 + x_1
   POL(true) = 2


The following pairs  are in P_>:
   IF(false, x, y, z) -> LOOP(x, double(y), s(z))
The following pairs are in P_bound:
   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)
The following rules are usable:
   true -> le(0, y)
   le(x, y) -> le(s(x), s(y))
   0 -> double(0)
   s(s(double(x))) -> double(s(x))
   false -> le(s(x), 0)

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

(27)
Complex Obligation (AND)

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

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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

(30)
TRUE

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

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

   IF(false, x, y, z) -> LOOP(x, double(y), s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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

(33)
TRUE