/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) AAECC Innermost [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) NonInfProof [EQUIVALENT, 12 ms]
        (20) QDP
        (21) QDPOrderProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) DependencyGraphProof [EQUIVALENT, 0 ms]
        (24) TRUE


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

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

   cond1(true, x, y) -> cond2(gr(x, y), x, y)
   cond2(true, x, y) -> cond1(gr(x, 0), y, y)
   cond2(false, x, y) -> cond1(gr(x, 0), p(x), y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   p(0) -> 0
   p(s(x)) -> x

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   p(0) -> 0
   p(s(x)) -> x

The TRS R 2 is 
   cond1(true, x, y) -> cond2(gr(x, y), x, y)
   cond2(true, x, y) -> cond1(gr(x, 0), y, y)
   cond2(false, x, y) -> cond1(gr(x, 0), p(x), y)

The signature Sigma is {cond1_3, cond2_3}
----------------------------------------

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

   cond1(true, x, y) -> cond2(gr(x, y), x, y)
   cond2(true, x, y) -> cond1(gr(x, 0), y, y)
   cond2(false, x, y) -> cond1(gr(x, 0), p(x), y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   p(0)
   p(s(x0))


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

(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:

   COND1(true, x, y) -> COND2(gr(x, y), x, y)
   COND1(true, x, y) -> GR(x, y)
   COND2(true, x, y) -> COND1(gr(x, 0), y, y)
   COND2(true, x, y) -> GR(x, 0)
   COND2(false, x, y) -> COND1(gr(x, 0), p(x), y)
   COND2(false, x, y) -> GR(x, 0)
   COND2(false, x, y) -> P(x)
   GR(s(x), s(y)) -> GR(x, y)

The TRS R consists of the following rules:

   cond1(true, x, y) -> cond2(gr(x, y), x, y)
   cond2(true, x, y) -> cond1(gr(x, 0), y, y)
   cond2(false, x, y) -> cond1(gr(x, 0), p(x), y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   p(0)
   p(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   cond1(true, x, y) -> cond2(gr(x, y), x, y)
   cond2(true, x, y) -> cond1(gr(x, 0), y, y)
   cond2(false, x, y) -> cond1(gr(x, 0), p(x), y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   p(0)
   p(s(x0))

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:

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

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

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   p(0)
   p(s(x0))

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].

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   p(0)
   p(s(x0))


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

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

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

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

   COND2(true, x, y) -> COND1(gr(x, 0), y, y)
   COND1(true, x, y) -> COND2(gr(x, y), x, y)
   COND2(false, x, y) -> COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

   cond1(true, x, y) -> cond2(gr(x, y), x, y)
   cond2(true, x, y) -> cond1(gr(x, 0), y, y)
   cond2(false, x, y) -> cond1(gr(x, 0), p(x), y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   p(0)
   p(s(x0))

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:

   COND2(true, x, y) -> COND1(gr(x, 0), y, y)
   COND1(true, x, y) -> COND2(gr(x, y), x, y)
   COND2(false, x, y) -> COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   p(0) -> 0
   p(s(x)) -> x
   gr(s(x), s(y)) -> gr(x, y)

The set Q consists of the following terms:

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   p(0)
   p(s(x0))

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].

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)


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

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

   COND2(true, x, y) -> COND1(gr(x, 0), y, y)
   COND1(true, x, y) -> COND2(gr(x, y), x, y)
   COND2(false, x, y) -> COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   p(0) -> 0
   p(s(x)) -> x
   gr(s(x), s(y)) -> gr(x, y)

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   p(0)
   p(s(x0))

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

(19) 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 COND2(true, x, y) -> COND1(gr(x, 0), y, y) the following chains were created:
*We consider the chain COND1(true, x2, x3) -> COND2(gr(x2, x3), x2, x3), COND2(true, x4, x5) -> COND1(gr(x4, 0), x5, x5) which results in the following constraint:

(1)    (COND2(gr(x2, x3), x2, x3)=COND2(true, x4, x5)  ==>  COND2(true, x4, x5)_>=_COND1(gr(x4, 0), x5, x5))



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

(2)    (gr(x2, x3)=true  ==>  COND2(true, x2, x3)_>=_COND1(gr(x2, 0), x3, x3))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x2, x3)=true which results in the following new constraints:

(3)    (true=true  ==>  COND2(true, s(x27), 0)_>=_COND1(gr(s(x27), 0), 0, 0))

(4)    (gr(x29, x28)=true & (gr(x29, x28)=true  ==>  COND2(true, x29, x28)_>=_COND1(gr(x29, 0), x28, x28))  ==>  COND2(true, s(x29), s(x28))_>=_COND1(gr(s(x29), 0), s(x28), s(x28)))



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

(5)    (COND2(true, s(x27), 0)_>=_COND1(gr(s(x27), 0), 0, 0))



We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gr(x29, x28)=true  ==>  COND2(true, x29, x28)_>=_COND1(gr(x29, 0), x28, x28)) with sigma = [ ] which results in the following new constraint:

(6)    (COND2(true, x29, x28)_>=_COND1(gr(x29, 0), x28, x28)  ==>  COND2(true, s(x29), s(x28))_>=_COND1(gr(s(x29), 0), s(x28), s(x28)))








For Pair COND1(true, x, y) -> COND2(gr(x, y), x, y) the following chains were created:
*We consider the chain COND2(true, x8, x9) -> COND1(gr(x8, 0), x9, x9), COND1(true, x10, x11) -> COND2(gr(x10, x11), x10, x11) which results in the following constraint:

(1)    (COND1(gr(x8, 0), x9, x9)=COND1(true, x10, x11)  ==>  COND1(true, x10, x11)_>=_COND2(gr(x10, x11), x10, x11))



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

(2)    (0=x30 & gr(x8, x30)=true  ==>  COND1(true, x9, x9)_>=_COND2(gr(x9, x9), x9, x9))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x8, x30)=true which results in the following new constraints:

(3)    (true=true & 0=0  ==>  COND1(true, x9, x9)_>=_COND2(gr(x9, x9), x9, x9))

(4)    (gr(x34, x33)=true & 0=s(x33) & (\/x35:gr(x34, x33)=true & 0=x33  ==>  COND1(true, x35, x35)_>=_COND2(gr(x35, x35), x35, x35))  ==>  COND1(true, x9, x9)_>=_COND2(gr(x9, x9), x9, x9))



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

(5)    (COND1(true, x9, x9)_>=_COND2(gr(x9, x9), x9, x9))



We solved constraint (4) using rules (I), (II).
*We consider the chain COND2(false, x14, x15) -> COND1(gr(x14, 0), p(x14), x15), COND1(true, x16, x17) -> COND2(gr(x16, x17), x16, x17) which results in the following constraint:

(1)    (COND1(gr(x14, 0), p(x14), x15)=COND1(true, x16, x17)  ==>  COND1(true, x16, x17)_>=_COND2(gr(x16, x17), x16, x17))



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

(2)    (0=x36 & gr(x14, x36)=true & p(x14)=x16  ==>  COND1(true, x16, x15)_>=_COND2(gr(x16, x15), x16, x15))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x14, x36)=true which results in the following new constraints:

(3)    (true=true & 0=0 & p(s(x38))=x16  ==>  COND1(true, x16, x15)_>=_COND2(gr(x16, x15), x16, x15))

(4)    (gr(x40, x39)=true & 0=s(x39) & p(s(x40))=x16 & (\/x41,x42:gr(x40, x39)=true & 0=x39 & p(x40)=x41  ==>  COND1(true, x41, x42)_>=_COND2(gr(x41, x42), x41, x42))  ==>  COND1(true, x16, x15)_>=_COND2(gr(x16, x15), x16, x15))



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

(5)    (COND1(true, x16, x15)_>=_COND2(gr(x16, x15), x16, x15))



We solved constraint (4) using rules (I), (II).




For Pair COND2(false, x, y) -> COND1(gr(x, 0), p(x), y) the following chains were created:
*We consider the chain COND1(true, x20, x21) -> COND2(gr(x20, x21), x20, x21), COND2(false, x22, x23) -> COND1(gr(x22, 0), p(x22), x23) which results in the following constraint:

(1)    (COND2(gr(x20, x21), x20, x21)=COND2(false, x22, x23)  ==>  COND2(false, x22, x23)_>=_COND1(gr(x22, 0), p(x22), x23))



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

(2)    (gr(x20, x21)=false  ==>  COND2(false, x20, x21)_>=_COND1(gr(x20, 0), p(x20), x21))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x20, x21)=false which results in the following new constraints:

(3)    (false=false  ==>  COND2(false, 0, x44)_>=_COND1(gr(0, 0), p(0), x44))

(4)    (gr(x47, x46)=false & (gr(x47, x46)=false  ==>  COND2(false, x47, x46)_>=_COND1(gr(x47, 0), p(x47), x46))  ==>  COND2(false, s(x47), s(x46))_>=_COND1(gr(s(x47), 0), p(s(x47)), s(x46)))



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

(5)    (COND2(false, 0, x44)_>=_COND1(gr(0, 0), p(0), x44))



We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gr(x47, x46)=false  ==>  COND2(false, x47, x46)_>=_COND1(gr(x47, 0), p(x47), x46)) with sigma = [ ] which results in the following new constraint:

(6)    (COND2(false, x47, x46)_>=_COND1(gr(x47, 0), p(x47), x46)  ==>  COND2(false, s(x47), s(x46))_>=_COND1(gr(s(x47), 0), p(s(x47)), s(x46)))








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

*COND2(true, x, y) -> COND1(gr(x, 0), y, y)

*(COND2(true, s(x27), 0)_>=_COND1(gr(s(x27), 0), 0, 0))


*(COND2(true, x29, x28)_>=_COND1(gr(x29, 0), x28, x28)  ==>  COND2(true, s(x29), s(x28))_>=_COND1(gr(s(x29), 0), s(x28), s(x28)))




*COND1(true, x, y) -> COND2(gr(x, y), x, y)

*(COND1(true, x9, x9)_>=_COND2(gr(x9, x9), x9, x9))


*(COND1(true, x16, x15)_>=_COND2(gr(x16, x15), x16, x15))




*COND2(false, x, y) -> COND1(gr(x, 0), p(x), y)

*(COND2(false, 0, x44)_>=_COND1(gr(0, 0), p(0), x44))


*(COND2(false, x47, x46)_>=_COND1(gr(x47, 0), p(x47), x46)  ==>  COND2(false, s(x47), s(x46))_>=_COND1(gr(s(x47), 0), p(s(x47)), s(x46)))








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(COND1(x_1, x_2, x_3)) = -1 + x_2 - x_3
   POL(COND2(x_1, x_2, x_3)) = -1 + x_1 + x_2 - x_3
   POL(c) = -1
   POL(false) = 0
   POL(gr(x_1, x_2)) = 0
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 1 + x_1
   POL(true) = 0


The following pairs  are in P_>:
   COND2(true, x, y) -> COND1(gr(x, 0), y, y)
The following pairs are in P_bound:
   COND2(true, x, y) -> COND1(gr(x, 0), y, y)
The following rules are usable:
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   p(0) -> 0
   p(s(x)) -> x

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

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

   COND1(true, x, y) -> COND2(gr(x, y), x, y)
   COND2(false, x, y) -> COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   p(0) -> 0
   p(s(x)) -> x
   gr(s(x), s(y)) -> gr(x, y)

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   p(0)
   p(s(x0))

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

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


The following pairs can be oriented strictly and are deleted.

   COND2(false, x, y) -> COND1(gr(x, 0), p(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(0) = 0
   POL(COND1(x_1, x_2, x_3)) = x_1 + x_2
   POL(COND2(x_1, x_2, x_3)) = [1] + [1/2]x_2
   POL(false) = 0
   POL(gr(x_1, x_2)) = [1/4]x_1
   POL(p(x_1)) = [1/4]x_1
   POL(s(x_1)) = [4] + [4]x_1
   POL(true) = [1]
The value of delta used in the strict ordering is 1.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   p(0) -> 0
   p(s(x)) -> x


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

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

   COND1(true, x, y) -> COND2(gr(x, y), x, y)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   p(0) -> 0
   p(s(x)) -> x
   gr(s(x), s(y)) -> gr(x, y)

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   p(0)
   p(s(x0))

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

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

(24)
TRUE