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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 25 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) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES


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

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

   f(f(x1)) -> b(b(b(x1)))
   a(f(x1)) -> f(a(a(x1)))
   b(b(x1)) -> c(c(a(c(x1))))
   d(b(x1)) -> d(a(b(x1)))
   c(c(x1)) -> d(d(d(x1)))
   b(d(x1)) -> d(b(x1))
   c(d(d(x1))) -> f(x1)

Q is empty.

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

(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

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

   f(f(x1)) -> b(b(b(x1)))
   f(a(x1)) -> a(a(f(x1)))
   b(b(x1)) -> c(a(c(c(x1))))
   b(d(x1)) -> b(a(d(x1)))
   c(c(x1)) -> d(d(d(x1)))
   d(b(x1)) -> b(d(x1))
   d(d(c(x1))) -> f(x1)

Q is empty.

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

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

   POL(a(x_1)) = x_1
   POL(b(x_1)) = 53 + x_1
   POL(c(x_1)) = 35 + x_1
   POL(d(x_1)) = 23 + x_1
   POL(f(x_1)) = 80 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   f(f(x1)) -> b(b(b(x1)))
   b(b(x1)) -> c(a(c(c(x1))))
   c(c(x1)) -> d(d(d(x1)))
   d(d(c(x1))) -> f(x1)




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

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

   f(a(x1)) -> a(a(f(x1)))
   b(d(x1)) -> b(a(d(x1)))
   d(b(x1)) -> b(d(x1))

Q is empty.

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

(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:

   F(a(x1)) -> F(x1)
   B(d(x1)) -> B(a(d(x1)))
   D(b(x1)) -> B(d(x1))
   D(b(x1)) -> D(x1)

The TRS R consists of the following rules:

   f(a(x1)) -> a(a(f(x1)))
   b(d(x1)) -> b(a(d(x1)))
   d(b(x1)) -> b(d(x1))

Q is empty.
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 2 less nodes.
----------------------------------------

(8)
Complex Obligation (AND)

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

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

   D(b(x1)) -> D(x1)

The TRS R consists of the following rules:

   f(a(x1)) -> a(a(f(x1)))
   b(d(x1)) -> b(a(d(x1)))
   d(b(x1)) -> b(d(x1))

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

(10) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   D(b(x1)) -> D(x1)

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


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

(13)
YES

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

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

   F(a(x1)) -> F(x1)

The TRS R consists of the following rules:

   f(a(x1)) -> a(a(f(x1)))
   b(d(x1)) -> b(a(d(x1)))
   d(b(x1)) -> b(d(x1))

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

(15) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   F(a(x1)) -> F(x1)

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

(17) 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(a(x1)) -> F(x1)
The graph contains the following edges 1 > 1


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

(18)
YES