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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) Overlay + Local Confluence [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, 2 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) NonInfProof [EQUIVALENT, 35 ms]
        (20) QDP
        (21) DependencyGraphProof [EQUIVALENT, 0 ms]
        (22) TRUE


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

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

   minus(x, y) -> cond(gt(x, y), x, y)
   cond(false, x, y) -> 0
   cond(true, x, y) -> s(minus(x, s(y)))
   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(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:

   minus(x, y) -> cond(gt(x, y), x, y)
   cond(false, x, y) -> 0
   cond(true, x, y) -> s(minus(x, s(y)))
   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(u, v)

The set Q consists of the following terms:

   minus(x0, x1)
   cond(false, x0, x1)
   cond(true, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(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:

   MINUS(x, y) -> COND(gt(x, y), x, y)
   MINUS(x, y) -> GT(x, y)
   COND(true, x, y) -> MINUS(x, s(y))
   GT(s(u), s(v)) -> GT(u, v)

The TRS R consists of the following rules:

   minus(x, y) -> cond(gt(x, y), x, y)
   cond(false, x, y) -> 0
   cond(true, x, y) -> s(minus(x, s(y)))
   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(u, v)

The set Q consists of the following terms:

   minus(x0, x1)
   cond(false, x0, x1)
   cond(true, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(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 2 SCCs with 1 less node.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   minus(x, y) -> cond(gt(x, y), x, y)
   cond(false, x, y) -> 0
   cond(true, x, y) -> s(minus(x, s(y)))
   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(u, v)

The set Q consists of the following terms:

   minus(x0, x1)
   cond(false, x0, x1)
   cond(true, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(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:

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

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

   minus(x0, x1)
   cond(false, x0, x1)
   cond(true, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(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].

   minus(x0, x1)
   cond(false, x0, x1)
   cond(true, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))


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

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

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

   COND(true, x, y) -> MINUS(x, s(y))
   MINUS(x, y) -> COND(gt(x, y), x, y)

The TRS R consists of the following rules:

   minus(x, y) -> cond(gt(x, y), x, y)
   cond(false, x, y) -> 0
   cond(true, x, y) -> s(minus(x, s(y)))
   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(u, v)

The set Q consists of the following terms:

   minus(x0, x1)
   cond(false, x0, x1)
   cond(true, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(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:

   COND(true, x, y) -> MINUS(x, s(y))
   MINUS(x, y) -> COND(gt(x, y), x, y)

The TRS R consists of the following rules:

   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(u, v)

The set Q consists of the following terms:

   minus(x0, x1)
   cond(false, x0, x1)
   cond(true, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(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].

   minus(x0, x1)
   cond(false, x0, x1)
   cond(true, x0, x1)


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

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

   COND(true, x, y) -> MINUS(x, s(y))
   MINUS(x, y) -> COND(gt(x, y), x, y)

The TRS R consists of the following rules:

   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(u, v)

The set Q consists of the following terms:

   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))

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 COND(true, x, y) -> MINUS(x, s(y)) the following chains were created:
*We consider the chain MINUS(x2, x3) -> COND(gt(x2, x3), x2, x3), COND(true, x4, x5) -> MINUS(x4, s(x5)) which results in the following constraint:

(1)    (COND(gt(x2, x3), x2, x3)=COND(true, x4, x5)  ==>  COND(true, x4, x5)_>=_MINUS(x4, s(x5)))



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

(2)    (gt(x2, x3)=true  ==>  COND(true, x2, x3)_>=_MINUS(x2, s(x3)))



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

(3)    (true=true  ==>  COND(true, s(x13), 0)_>=_MINUS(s(x13), s(0)))

(4)    (gt(x15, x14)=true & (gt(x15, x14)=true  ==>  COND(true, x15, x14)_>=_MINUS(x15, s(x14)))  ==>  COND(true, s(x15), s(x14))_>=_MINUS(s(x15), s(s(x14))))



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

(5)    (COND(true, s(x13), 0)_>=_MINUS(s(x13), s(0)))



We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gt(x15, x14)=true  ==>  COND(true, x15, x14)_>=_MINUS(x15, s(x14))) with sigma = [ ] which results in the following new constraint:

(6)    (COND(true, x15, x14)_>=_MINUS(x15, s(x14))  ==>  COND(true, s(x15), s(x14))_>=_MINUS(s(x15), s(s(x14))))








For Pair MINUS(x, y) -> COND(gt(x, y), x, y) the following chains were created:
*We consider the chain COND(true, x6, x7) -> MINUS(x6, s(x7)), MINUS(x8, x9) -> COND(gt(x8, x9), x8, x9) which results in the following constraint:

(1)    (MINUS(x6, s(x7))=MINUS(x8, x9)  ==>  MINUS(x8, x9)_>=_COND(gt(x8, x9), x8, x9))



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

(2)    (MINUS(x6, s(x7))_>=_COND(gt(x6, s(x7)), x6, s(x7)))








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

*COND(true, x, y) -> MINUS(x, s(y))

*(COND(true, s(x13), 0)_>=_MINUS(s(x13), s(0)))


*(COND(true, x15, x14)_>=_MINUS(x15, s(x14))  ==>  COND(true, s(x15), s(x14))_>=_MINUS(s(x15), s(s(x14))))




*MINUS(x, y) -> COND(gt(x, y), x, y)

*(MINUS(x6, s(x7))_>=_COND(gt(x6, s(x7)), x6, s(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) = 1
   POL(COND(x_1, x_2, x_3)) = -1 - x_1 + x_2 - x_3
   POL(MINUS(x_1, x_2)) = -1 + x_1 - x_2
   POL(c) = -2
   POL(false) = 0
   POL(gt(x_1, x_2)) = 0
   POL(s(x_1)) = 1 + x_1
   POL(true) = 0


The following pairs  are in P_>:
   COND(true, x, y) -> MINUS(x, s(y))
The following pairs are in P_bound:
   COND(true, x, y) -> MINUS(x, s(y))
The following rules are usable:
   false -> gt(0, v)
   true -> gt(s(u), 0)
   gt(u, v) -> gt(s(u), s(v))

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

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

   MINUS(x, y) -> COND(gt(x, y), x, y)

The TRS R consists of the following rules:

   gt(0, v) -> false
   gt(s(u), 0) -> true
   gt(s(u), s(v)) -> gt(u, v)

The set Q consists of the following terms:

   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))

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

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

(22)
TRUE