/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 14 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


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

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

   minus(0, Y) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   geq(X, 0) -> true
   geq(0, s(Y)) -> false
   geq(s(X), s(Y)) -> geq(X, Y)
   div(0, s(Y)) -> 0
   div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   if(true, X, Y) -> X
   if(false, X, Y) -> Y

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(0, Y) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   geq(X, 0) -> true
   geq(0, s(Y)) -> false
   geq(s(X), s(Y)) -> geq(X, Y)
   div(0, s(Y)) -> 0
   div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   if(true, X, Y) -> X
   if(false, X, Y) -> Y

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), s(x1))
   geq(x0, 0)
   geq(0, s(x0))
   geq(s(x0), s(x1))
   div(0, s(x0))
   div(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, 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(s(X), s(Y)) -> MINUS(X, Y)
   GEQ(s(X), s(Y)) -> GEQ(X, Y)
   DIV(s(X), s(Y)) -> IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   DIV(s(X), s(Y)) -> GEQ(X, Y)
   DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))
   DIV(s(X), s(Y)) -> MINUS(X, Y)

The TRS R consists of the following rules:

   minus(0, Y) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   geq(X, 0) -> true
   geq(0, s(Y)) -> false
   geq(s(X), s(Y)) -> geq(X, Y)
   div(0, s(Y)) -> 0
   div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   if(true, X, Y) -> X
   if(false, X, Y) -> Y

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), s(x1))
   geq(x0, 0)
   geq(0, s(x0))
   geq(s(x0), s(x1))
   div(0, s(x0))
   div(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, 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 4 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   GEQ(s(X), s(Y)) -> GEQ(X, Y)

The TRS R consists of the following rules:

   minus(0, Y) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   geq(X, 0) -> true
   geq(0, s(Y)) -> false
   geq(s(X), s(Y)) -> geq(X, Y)
   div(0, s(Y)) -> 0
   div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   if(true, X, Y) -> X
   if(false, X, Y) -> Y

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), s(x1))
   geq(x0, 0)
   geq(0, s(x0))
   geq(s(x0), s(x1))
   div(0, s(x0))
   div(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, 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:

   GEQ(s(X), s(Y)) -> GEQ(X, Y)

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

   minus(0, x0)
   minus(s(x0), s(x1))
   geq(x0, 0)
   geq(0, s(x0))
   geq(s(x0), s(x1))
   div(0, s(x0))
   div(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, 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(0, x0)
   minus(s(x0), s(x1))
   geq(x0, 0)
   geq(0, s(x0))
   geq(s(x0), s(x1))
   div(0, s(x0))
   div(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)


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

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

   GEQ(s(X), s(Y)) -> GEQ(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:
*GEQ(s(X), s(Y)) -> GEQ(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:

   MINUS(s(X), s(Y)) -> MINUS(X, Y)

The TRS R consists of the following rules:

   minus(0, Y) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   geq(X, 0) -> true
   geq(0, s(Y)) -> false
   geq(s(X), s(Y)) -> geq(X, Y)
   div(0, s(Y)) -> 0
   div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
   if(true, X, Y) -> X
   if(false, X, Y) -> Y

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), s(x1))
   geq(x0, 0)
   geq(0, s(x0))
   geq(s(x0), s(x1))
   div(0, s(x0))
   div(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, 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:

   MINUS(s(X), s(Y)) -> MINUS(X, Y)

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

   minus(0, x0)
   minus(s(x0), s(x1))
   geq(x0, 0)
   geq(0, s(x0))
   geq(s(x0), s(x1))
   div(0, s(x0))
   div(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, 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(0, x0)
   minus(s(x0), s(x1))
   geq(x0, 0)
   geq(0, s(x0))
   geq(s(x0), s(x1))
   div(0, s(x0))
   div(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)


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

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

(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:
*MINUS(s(X), s(Y)) -> MINUS(X, Y)
The graph contains the following edges 1 > 1, 2 > 2


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

(20)
YES