YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QReductionProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QReductionProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) TransformationProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QReductionProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) TransformationProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) DependencyGraphProof [EQUIVALENT, 0 ms]
        (26) TRUE


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

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

   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))
   double(X) -> +(X, X)
   f(0, s(0), X) -> f(X, double(X), X)
   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   double(x0)
   f(0, s(0), x0)
   g(x0, x1)


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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   +^1(X, s(Y)) -> +^1(X, Y)
   DOUBLE(X) -> +^1(X, X)
   F(0, s(0), X) -> F(X, double(X), X)
   F(0, s(0), X) -> DOUBLE(X)

The TRS R consists of the following rules:

   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))
   double(X) -> +(X, X)
   f(0, s(0), X) -> f(X, double(X), X)
   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   double(x0)
   f(0, s(0), x0)
   g(x0, x1)

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

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

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   +^1(X, s(Y)) -> +^1(X, Y)

The TRS R consists of the following rules:

   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))
   double(X) -> +(X, X)
   f(0, s(0), X) -> f(X, double(X), X)
   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   double(x0)
   f(0, s(0), x0)
   g(x0, x1)

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

(6) 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.
----------------------------------------

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

   +^1(X, s(Y)) -> +^1(X, Y)

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

   +(x0, 0)
   +(x0, s(x1))
   double(x0)
   f(0, s(0), x0)
   g(x0, x1)

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

(8) 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].

   +(x0, 0)
   +(x0, s(x1))
   double(x0)
   f(0, s(0), x0)
   g(x0, x1)


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

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   +^1(X, s(Y)) -> +^1(X, Y)

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

(10) 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:
*+^1(X, s(Y)) -> +^1(X, Y)
The graph contains the following edges 1 >= 1, 2 > 2


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

(11)
YES

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

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(0, s(0), X) -> F(X, double(X), X)

The TRS R consists of the following rules:

   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))
   double(X) -> +(X, X)
   f(0, s(0), X) -> f(X, double(X), X)
   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   double(x0)
   f(0, s(0), x0)
   g(x0, x1)

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

(13) 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.
----------------------------------------

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(0, s(0), X) -> F(X, double(X), X)

The TRS R consists of the following rules:

   double(X) -> +(X, X)
   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   double(x0)
   f(0, s(0), x0)
   g(x0, x1)

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

(15) 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].

   f(0, s(0), x0)
   g(x0, x1)


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

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(0, s(0), X) -> F(X, double(X), X)

The TRS R consists of the following rules:

   double(X) -> +(X, X)
   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))

The set Q consists of the following terms:

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

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

(17) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule F(0, s(0), X) -> F(X, double(X), X) at position [1] we obtained the following new rules [LPAR04]:

   (F(0, s(0), X) -> F(X, +(X, X), X),F(0, s(0), X) -> F(X, +(X, X), X))


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

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

   F(0, s(0), X) -> F(X, +(X, X), X)

The TRS R consists of the following rules:

   double(X) -> +(X, X)
   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))

The set Q consists of the following terms:

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

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

(19) 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.
----------------------------------------

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

   F(0, s(0), X) -> F(X, +(X, X), X)

The TRS R consists of the following rules:

   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))

The set Q consists of the following terms:

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

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

(21) 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].

   double(x0)


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

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

   F(0, s(0), X) -> F(X, +(X, X), X)

The TRS R consists of the following rules:

   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))

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

(23) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule F(0, s(0), X) -> F(X, +(X, X), X) at position [1] we obtained the following new rules [LPAR04]:

   (F(0, s(0), 0) -> F(0, 0, 0),F(0, s(0), 0) -> F(0, 0, 0))
   (F(0, s(0), s(x1)) -> F(s(x1), s(+(s(x1), x1)), s(x1)),F(0, s(0), s(x1)) -> F(s(x1), s(+(s(x1), x1)), s(x1)))


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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(0, s(0), 0) -> F(0, 0, 0)
   F(0, s(0), s(x1)) -> F(s(x1), s(+(s(x1), x1)), s(x1))

The TRS R consists of the following rules:

   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))

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

(25) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
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(26)
TRUE