/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 41 ms]
(2) QTRS
(3) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 18 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 3 ms]
(8) AND
    (9) QDP
        (10) UsableRulesProof [EQUIVALENT, 1 ms]
        (11) QDP
        (12) QReductionProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) MRRProof [EQUIVALENT, 4 ms]
        (15) QDP
        (16) DependencyGraphProof [EQUIVALENT, 0 ms]
        (17) TRUE
    (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QReductionProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) MRRProof [EQUIVALENT, 4 ms]
        (24) QDP
        (25) DependencyGraphProof [EQUIVALENT, 0 ms]
        (26) TRUE


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

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

   f(0(x1)) -> s(0(x1))
   d(0(x1)) -> 0(x1)
   d(s(x1)) -> s(s(d(p(s(x1)))))
   f(s(x1)) -> d(f(p(s(x1))))
   p(s(x1)) -> x1

Q is empty.

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

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

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

   f(0(x1)) -> s(0(x1))




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

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

   d(0(x1)) -> 0(x1)
   d(s(x1)) -> s(s(d(p(s(x1)))))
   f(s(x1)) -> d(f(p(s(x1))))
   p(s(x1)) -> x1

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:

   d(0(x1)) -> 0(x1)
   d(s(x1)) -> s(s(d(p(s(x1)))))
   f(s(x1)) -> d(f(p(s(x1))))
   p(s(x1)) -> x1

The set Q consists of the following terms:

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

   D(s(x1)) -> D(p(s(x1)))
   D(s(x1)) -> P(s(x1))
   F(s(x1)) -> D(f(p(s(x1))))
   F(s(x1)) -> F(p(s(x1)))
   F(s(x1)) -> P(s(x1))

The TRS R consists of the following rules:

   d(0(x1)) -> 0(x1)
   d(s(x1)) -> s(s(d(p(s(x1)))))
   f(s(x1)) -> d(f(p(s(x1))))
   p(s(x1)) -> x1

The set Q consists of the following terms:

   d(0(x0))
   d(s(x0))
   f(s(x0))
   p(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 3 less nodes.
----------------------------------------

(8)
Complex Obligation (AND)

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

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

   D(s(x1)) -> D(p(s(x1)))

The TRS R consists of the following rules:

   d(0(x1)) -> 0(x1)
   d(s(x1)) -> s(s(d(p(s(x1)))))
   f(s(x1)) -> d(f(p(s(x1))))
   p(s(x1)) -> x1

The set Q consists of the following terms:

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

   D(s(x1)) -> D(p(s(x1)))

The TRS R consists of the following rules:

   p(s(x1)) -> x1

The set Q consists of the following terms:

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

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

(12) 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(x0))
   d(s(x0))
   f(s(x0))


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

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

   D(s(x1)) -> D(p(s(x1)))

The TRS R consists of the following rules:

   p(s(x1)) -> x1

The set Q consists of the following terms:

   p(s(x0))

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

(14) 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(x1)) -> x1

Used ordering: Polynomial interpretation [POLO]:

   POL(D(x_1)) = x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 1 + x_1


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

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

   D(s(x1)) -> D(p(s(x1)))

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

   p(s(x0))

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

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

(17)
TRUE

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

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

   F(s(x1)) -> F(p(s(x1)))

The TRS R consists of the following rules:

   d(0(x1)) -> 0(x1)
   d(s(x1)) -> s(s(d(p(s(x1)))))
   f(s(x1)) -> d(f(p(s(x1))))
   p(s(x1)) -> x1

The set Q consists of the following terms:

   d(0(x0))
   d(s(x0))
   f(s(x0))
   p(s(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(s(x1)) -> F(p(s(x1)))

The TRS R consists of the following rules:

   p(s(x1)) -> x1

The set Q consists of the following terms:

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

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


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

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

   F(s(x1)) -> F(p(s(x1)))

The TRS R consists of the following rules:

   p(s(x1)) -> x1

The set Q consists of the following terms:

   p(s(x0))

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

(23) 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(x1)) -> x1

Used ordering: Polynomial interpretation [POLO]:

   POL(F(x_1)) = x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 1 + x_1


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

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

   F(s(x1)) -> F(p(s(x1)))

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

   p(s(x0))

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 1 less node.
----------------------------------------

(26)
TRUE