/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 11 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) 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, 34 ms]
        (27) QDP
        (28) DependencyGraphProof [EQUIVALENT, 0 ms]
        (29) TRUE


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

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

   nthtail(n, l) -> cond(ge(n, length(l)), n, l)
   cond(true, n, l) -> l
   cond(false, n, l) -> tail(nthtail(s(n), l))
   tail(nil) -> nil
   tail(cons(x, l)) -> l
   length(nil) -> 0
   length(cons(x, l)) -> s(length(l))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

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:

   nthtail(n, l) -> cond(ge(n, length(l)), n, l)
   cond(true, n, l) -> l
   cond(false, n, l) -> tail(nthtail(s(n), l))
   tail(nil) -> nil
   tail(cons(x, l)) -> l
   length(nil) -> 0
   length(cons(x, l)) -> s(length(l))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

The set Q consists of the following terms:

   nthtail(x0, x1)
   cond(true, x0, x1)
   cond(false, x0, x1)
   tail(nil)
   tail(cons(x0, x1))
   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))


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

(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:

   NTHTAIL(n, l) -> COND(ge(n, length(l)), n, l)
   NTHTAIL(n, l) -> GE(n, length(l))
   NTHTAIL(n, l) -> LENGTH(l)
   COND(false, n, l) -> TAIL(nthtail(s(n), l))
   COND(false, n, l) -> NTHTAIL(s(n), l)
   LENGTH(cons(x, l)) -> LENGTH(l)
   GE(s(u), s(v)) -> GE(u, v)

The TRS R consists of the following rules:

   nthtail(n, l) -> cond(ge(n, length(l)), n, l)
   cond(true, n, l) -> l
   cond(false, n, l) -> tail(nthtail(s(n), l))
   tail(nil) -> nil
   tail(cons(x, l)) -> l
   length(nil) -> 0
   length(cons(x, l)) -> s(length(l))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

The set Q consists of the following terms:

   nthtail(x0, x1)
   cond(true, x0, x1)
   cond(false, x0, x1)
   tail(nil)
   tail(cons(x0, x1))
   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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:

   GE(s(u), s(v)) -> GE(u, v)

The TRS R consists of the following rules:

   nthtail(n, l) -> cond(ge(n, length(l)), n, l)
   cond(true, n, l) -> l
   cond(false, n, l) -> tail(nthtail(s(n), l))
   tail(nil) -> nil
   tail(cons(x, l)) -> l
   length(nil) -> 0
   length(cons(x, l)) -> s(length(l))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

The set Q consists of the following terms:

   nthtail(x0, x1)
   cond(true, x0, x1)
   cond(false, x0, x1)
   tail(nil)
   tail(cons(x0, x1))
   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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:

   GE(s(u), s(v)) -> GE(u, v)

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

   nthtail(x0, x1)
   cond(true, x0, x1)
   cond(false, x0, x1)
   tail(nil)
   tail(cons(x0, x1))
   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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].

   nthtail(x0, x1)
   cond(true, x0, x1)
   cond(false, x0, x1)
   tail(nil)
   tail(cons(x0, x1))
   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))


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

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

   GE(s(u), s(v)) -> GE(u, v)

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:
*GE(s(u), s(v)) -> GE(u, v)
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:

   LENGTH(cons(x, l)) -> LENGTH(l)

The TRS R consists of the following rules:

   nthtail(n, l) -> cond(ge(n, length(l)), n, l)
   cond(true, n, l) -> l
   cond(false, n, l) -> tail(nthtail(s(n), l))
   tail(nil) -> nil
   tail(cons(x, l)) -> l
   length(nil) -> 0
   length(cons(x, l)) -> s(length(l))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

The set Q consists of the following terms:

   nthtail(x0, x1)
   cond(true, x0, x1)
   cond(false, x0, x1)
   tail(nil)
   tail(cons(x0, x1))
   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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:

   LENGTH(cons(x, l)) -> LENGTH(l)

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

   nthtail(x0, x1)
   cond(true, x0, x1)
   cond(false, x0, x1)
   tail(nil)
   tail(cons(x0, x1))
   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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].

   nthtail(x0, x1)
   cond(true, x0, x1)
   cond(false, x0, x1)
   tail(nil)
   tail(cons(x0, x1))
   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))


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

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

   LENGTH(cons(x, l)) -> LENGTH(l)

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:
*LENGTH(cons(x, l)) -> LENGTH(l)
The graph contains the following edges 1 > 1


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

(20)
YES

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

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

   COND(false, n, l) -> NTHTAIL(s(n), l)
   NTHTAIL(n, l) -> COND(ge(n, length(l)), n, l)

The TRS R consists of the following rules:

   nthtail(n, l) -> cond(ge(n, length(l)), n, l)
   cond(true, n, l) -> l
   cond(false, n, l) -> tail(nthtail(s(n), l))
   tail(nil) -> nil
   tail(cons(x, l)) -> l
   length(nil) -> 0
   length(cons(x, l)) -> s(length(l))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

The set Q consists of the following terms:

   nthtail(x0, x1)
   cond(true, x0, x1)
   cond(false, x0, x1)
   tail(nil)
   tail(cons(x0, x1))
   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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:

   COND(false, n, l) -> NTHTAIL(s(n), l)
   NTHTAIL(n, l) -> COND(ge(n, length(l)), n, l)

The TRS R consists of the following rules:

   length(nil) -> 0
   length(cons(x, l)) -> s(length(l))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

The set Q consists of the following terms:

   nthtail(x0, x1)
   cond(true, x0, x1)
   cond(false, x0, x1)
   tail(nil)
   tail(cons(x0, x1))
   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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].

   nthtail(x0, x1)
   cond(true, x0, x1)
   cond(false, x0, x1)
   tail(nil)
   tail(cons(x0, x1))


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

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

   COND(false, n, l) -> NTHTAIL(s(n), l)
   NTHTAIL(n, l) -> COND(ge(n, length(l)), n, l)

The TRS R consists of the following rules:

   length(nil) -> 0
   length(cons(x, l)) -> s(length(l))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

The set Q consists of the following terms:

   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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 COND(false, n, l) -> NTHTAIL(s(n), l) the following chains were created:
*We consider the chain NTHTAIL(x2, x3) -> COND(ge(x2, length(x3)), x2, x3), COND(false, x4, x5) -> NTHTAIL(s(x4), x5) which results in the following constraint:

(1)    (COND(ge(x2, length(x3)), x2, x3)=COND(false, x4, x5)  ==>  COND(false, x4, x5)_>=_NTHTAIL(s(x4), x5))



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

(2)    (length(x3)=x12 & ge(x2, x12)=false  ==>  COND(false, x2, x3)_>=_NTHTAIL(s(x2), x3))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x2, x12)=false which results in the following new constraints:

(3)    (false=false & length(x3)=s(x14)  ==>  COND(false, 0, x3)_>=_NTHTAIL(s(0), x3))

(4)    (ge(x16, x15)=false & length(x3)=s(x15) & (\/x17:ge(x16, x15)=false & length(x17)=x15  ==>  COND(false, x16, x17)_>=_NTHTAIL(s(x16), x17))  ==>  COND(false, s(x16), x3)_>=_NTHTAIL(s(s(x16)), x3))



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

(5)    (length(x3)=s(x14)  ==>  COND(false, 0, x3)_>=_NTHTAIL(s(0), x3))



We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on length(x3)=s(x15) which results in the following new constraint:

(6)    (s(length(x21))=s(x15) & ge(x16, x15)=false & (\/x17:ge(x16, x15)=false & length(x17)=x15  ==>  COND(false, x16, x17)_>=_NTHTAIL(s(x16), x17)) & (\/x23,x24,x25:length(x21)=s(x23) & ge(x24, x23)=false & (\/x25:ge(x24, x23)=false & length(x25)=x23  ==>  COND(false, x24, x25)_>=_NTHTAIL(s(x24), x25))  ==>  COND(false, s(x24), x21)_>=_NTHTAIL(s(s(x24)), x21))  ==>  COND(false, s(x16), cons(x22, x21))_>=_NTHTAIL(s(s(x16)), cons(x22, x21)))



We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on length(x3)=s(x14) which results in the following new constraint:

(7)    (s(length(x18))=s(x14) & (\/x20:length(x18)=s(x20)  ==>  COND(false, 0, x18)_>=_NTHTAIL(s(0), x18))  ==>  COND(false, 0, cons(x19, x18))_>=_NTHTAIL(s(0), cons(x19, x18)))



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

(8)    (COND(false, 0, cons(x19, x18))_>=_NTHTAIL(s(0), cons(x19, x18)))



We simplified constraint (6) using rules (I), (II) which results in the following new constraint:

(9)    (length(x21)=x15 & ge(x16, x15)=false & (\/x17:ge(x16, x15)=false & length(x17)=x15  ==>  COND(false, x16, x17)_>=_NTHTAIL(s(x16), x17)) & (\/x23,x24,x25:length(x21)=s(x23) & ge(x24, x23)=false & (\/x25:ge(x24, x23)=false & length(x25)=x23  ==>  COND(false, x24, x25)_>=_NTHTAIL(s(x24), x25))  ==>  COND(false, s(x24), x21)_>=_NTHTAIL(s(s(x24)), x21))  ==>  COND(false, s(x16), cons(x22, x21))_>=_NTHTAIL(s(s(x16)), cons(x22, x21)))



We simplified constraint (9) using rule (VI) where we applied the induction hypothesis (\/x17:ge(x16, x15)=false & length(x17)=x15  ==>  COND(false, x16, x17)_>=_NTHTAIL(s(x16), x17)) with sigma = [x17 / x21] which results in the following new constraint:

(10)    (COND(false, x16, x21)_>=_NTHTAIL(s(x16), x21) & (\/x23,x24,x25:length(x21)=s(x23) & ge(x24, x23)=false & (\/x25:ge(x24, x23)=false & length(x25)=x23  ==>  COND(false, x24, x25)_>=_NTHTAIL(s(x24), x25))  ==>  COND(false, s(x24), x21)_>=_NTHTAIL(s(s(x24)), x21))  ==>  COND(false, s(x16), cons(x22, x21))_>=_NTHTAIL(s(s(x16)), cons(x22, x21)))



We simplified constraint (10) using rule (IV) which results in the following new constraint:

(11)    (COND(false, x16, x21)_>=_NTHTAIL(s(x16), x21)  ==>  COND(false, s(x16), cons(x22, x21))_>=_NTHTAIL(s(s(x16)), cons(x22, x21)))








For Pair NTHTAIL(n, l) -> COND(ge(n, length(l)), n, l) the following chains were created:
*We consider the chain COND(false, x6, x7) -> NTHTAIL(s(x6), x7), NTHTAIL(x8, x9) -> COND(ge(x8, length(x9)), x8, x9) which results in the following constraint:

(1)    (NTHTAIL(s(x6), x7)=NTHTAIL(x8, x9)  ==>  NTHTAIL(x8, x9)_>=_COND(ge(x8, length(x9)), x8, x9))



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

(2)    (NTHTAIL(s(x6), x7)_>=_COND(ge(s(x6), length(x7)), s(x6), x7))








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

*COND(false, n, l) -> NTHTAIL(s(n), l)

*(COND(false, 0, cons(x19, x18))_>=_NTHTAIL(s(0), cons(x19, x18)))


*(COND(false, x16, x21)_>=_NTHTAIL(s(x16), x21)  ==>  COND(false, s(x16), cons(x22, x21))_>=_NTHTAIL(s(s(x16)), cons(x22, x21)))




*NTHTAIL(n, l) -> COND(ge(n, length(l)), n, l)

*(NTHTAIL(s(x6), x7)_>=_COND(ge(s(x6), length(x7)), s(x6), x7))








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(COND(x_1, x_2, x_3)) = -1 + x_1 - x_2 + x_3
   POL(NTHTAIL(x_1, x_2)) = -1 - x_1 + x_2
   POL(c) = -2
   POL(cons(x_1, x_2)) = 1 + x_1 + x_2
   POL(false) = 0
   POL(ge(x_1, x_2)) = 0
   POL(length(x_1)) = 0
   POL(nil) = 1
   POL(s(x_1)) = 1 + x_1
   POL(true) = 0


The following pairs  are in P_>:
   COND(false, n, l) -> NTHTAIL(s(n), l)
The following pairs are in P_bound:
   COND(false, n, l) -> NTHTAIL(s(n), l)
The following rules are usable:
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

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

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

   NTHTAIL(n, l) -> COND(ge(n, length(l)), n, l)

The TRS R consists of the following rules:

   length(nil) -> 0
   length(cons(x, l)) -> s(length(l))
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)

The set Q consists of the following terms:

   length(nil)
   length(cons(x0, x1))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

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

(29)
TRUE