/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 88 ms]
(2) QTRS
(3) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) UsableRulesProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) ATransformationProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QReductionProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) MRRProof [EQUIVALENT, 5 ms]
        (17) QDP
        (18) DependencyGraphProof [EQUIVALENT, 0 ms]
        (19) TRUE
    (20) QDP
        (21) UsableRulesProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) ATransformationProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) QReductionProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) MRRProof [EQUIVALENT, 6 ms]
        (28) QDP
        (29) DependencyGraphProof [EQUIVALENT, 0 ms]
        (30) TRUE


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

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

   a(f, 0) -> a(s, 0)
   a(d, 0) -> 0
   a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
   a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
   a(p, a(s, x)) -> x

Q is empty.

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

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(a(x_1, x_2)) = 2*x_1 + x_2
   POL(d) = 0
   POL(f) = 1
   POL(p) = 0
   POL(s) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   a(f, 0) -> a(s, 0)




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

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

   a(d, 0) -> 0
   a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
   a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
   a(p, a(s, x)) -> x

Q is empty.

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

(3) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a(d, 0) -> 0
   a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
   a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
   a(p, a(s, x)) -> x

The set Q consists of the following terms:

   a(d, 0)
   a(d, a(s, x0))
   a(f, a(s, x0))
   a(p, a(s, x0))


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

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

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(d, a(s, x)) -> A(s, a(s, a(d, a(p, a(s, x)))))
   A(d, a(s, x)) -> A(s, a(d, a(p, a(s, x))))
   A(d, a(s, x)) -> A(d, a(p, a(s, x)))
   A(d, a(s, x)) -> A(p, a(s, x))
   A(f, a(s, x)) -> A(d, a(f, a(p, a(s, x))))
   A(f, a(s, x)) -> A(f, a(p, a(s, x)))
   A(f, a(s, x)) -> A(p, a(s, x))

The TRS R consists of the following rules:

   a(d, 0) -> 0
   a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
   a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
   a(p, a(s, x)) -> x

The set Q consists of the following terms:

   a(d, 0)
   a(d, a(s, x0))
   a(f, a(s, x0))
   a(p, a(s, x0))

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

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

(8)
Complex Obligation (AND)

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

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

   A(d, a(s, x)) -> A(d, a(p, a(s, x)))

The TRS R consists of the following rules:

   a(d, 0) -> 0
   a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
   a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
   a(p, a(s, x)) -> x

The set Q consists of the following terms:

   a(d, 0)
   a(d, a(s, x0))
   a(f, a(s, x0))
   a(p, a(s, x0))

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

(10) 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.
----------------------------------------

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

   A(d, a(s, x)) -> A(d, a(p, a(s, x)))

The TRS R consists of the following rules:

   a(p, a(s, x)) -> x

The set Q consists of the following terms:

   a(d, 0)
   a(d, a(s, x0))
   a(f, a(s, x0))
   a(p, a(s, x0))

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

(12) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   d1(s(x)) -> d1(p(s(x)))

The TRS R consists of the following rules:

   p(s(x)) -> x

The set Q consists of the following terms:

   d(0)
   d(s(x0))
   f(s(x0))
   p(s(x0))

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

(14) 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].

   d(0)
   d(s(x0))
   f(s(x0))


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

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   d1(s(x)) -> d1(p(s(x)))

The TRS R consists of the following rules:

   p(s(x)) -> x

The set Q consists of the following terms:

   p(s(x0))

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

(16) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   p(s(x)) -> x

Used ordering: Polynomial interpretation [POLO]:

   POL(d1(x_1)) = 2*x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 2 + x_1


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

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   d1(s(x)) -> d1(p(s(x)))

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

   p(s(x0))

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

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

(19)
TRUE

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

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

   A(f, a(s, x)) -> A(f, a(p, a(s, x)))

The TRS R consists of the following rules:

   a(d, 0) -> 0
   a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
   a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
   a(p, a(s, x)) -> x

The set Q consists of the following terms:

   a(d, 0)
   a(d, a(s, x0))
   a(f, a(s, x0))
   a(p, a(s, x0))

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

(21) 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.
----------------------------------------

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

   A(f, a(s, x)) -> A(f, a(p, a(s, x)))

The TRS R consists of the following rules:

   a(p, a(s, x)) -> x

The set Q consists of the following terms:

   a(d, 0)
   a(d, a(s, x0))
   a(f, a(s, x0))
   a(p, a(s, x0))

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

(23) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

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

   f1(s(x)) -> f1(p(s(x)))

The TRS R consists of the following rules:

   p(s(x)) -> x

The set Q consists of the following terms:

   d(0)
   d(s(x0))
   f(s(x0))
   p(s(x0))

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

(25) 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].

   d(0)
   d(s(x0))
   f(s(x0))


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

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   f1(s(x)) -> f1(p(s(x)))

The TRS R consists of the following rules:

   p(s(x)) -> x

The set Q consists of the following terms:

   p(s(x0))

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

(27) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   p(s(x)) -> x

Used ordering: Polynomial interpretation [POLO]:

   POL(f1(x_1)) = 2*x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 2 + x_1


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

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   f1(s(x)) -> f1(p(s(x)))

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

   p(s(x0))

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

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

(30)
TRUE