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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 29 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) QReductionProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QReductionProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (22) YES


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

The set Q consists of the following terms:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   d(0(x0))
   d(s(x0))
   f(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 1 less node.
----------------------------------------

(8)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   d(0(x0))
   d(s(x0))
   f(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(x1)

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

   d(0(x0))
   d(s(x0))
   f(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(x1)

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

(14) 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:
*D(s(x1)) -> D(x1)
The graph contains the following edges 1 > 1


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

(15)
YES

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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(17) 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.
----------------------------------------

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

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

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

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

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

(19) 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))


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

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

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

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

(21) 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:
*F(s(x1)) -> F(x1)
The graph contains the following edges 1 > 1


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

(22)
YES