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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 15 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPOrderProof [EQUIVALENT, 31 ms]
        (9) QDP
        (10) DependencyGraphProof [EQUIVALENT, 0 ms]
        (11) TRUE
    (12) QDP
        (13) MNOCProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPOrderProof [EQUIVALENT, 35 ms]
        (20) QDP
        (21) PisEmptyProof [EQUIVALENT, 0 ms]
        (22) YES


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

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

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   L(a(a(x1))) -> L(a(b(c(x1))))
   c(R(x1)) -> b(a(R(x1)))

Q is empty.

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(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:

   B(a(a(x1))) -> B(c(x1))
   B(a(a(x1))) -> C(x1)
   C(a(x1)) -> C(x1)
   C(b(x1)) -> B(a(x1))
   L^1(a(a(x1))) -> L^1(a(b(c(x1))))
   L^1(a(a(x1))) -> B(c(x1))
   L^1(a(a(x1))) -> C(x1)
   C(R(x1)) -> B(a(R(x1)))

The TRS R consists of the following rules:

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   L(a(a(x1))) -> L(a(b(c(x1))))
   c(R(x1)) -> b(a(R(x1)))

Q is empty.
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 3 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   B(a(a(x1))) -> C(x1)
   C(a(x1)) -> C(x1)
   C(b(x1)) -> B(a(x1))
   B(a(a(x1))) -> B(c(x1))

The TRS R consists of the following rules:

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   L(a(a(x1))) -> L(a(b(c(x1))))
   c(R(x1)) -> b(a(R(x1)))

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

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

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

   B(a(a(x1))) -> C(x1)
   C(a(x1)) -> C(x1)
   C(b(x1)) -> B(a(x1))
   B(a(a(x1))) -> B(c(x1))

The TRS R consists of the following rules:

   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   c(R(x1)) -> b(a(R(x1)))
   b(a(a(x1))) -> a(b(c(x1)))

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

(8) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   B(a(a(x1))) -> C(x1)
   C(a(x1)) -> C(x1)
   B(a(a(x1))) -> B(c(x1))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(B(x_1)) = x_1
   POL(C(x_1)) = 1 + x_1
   POL(R(x_1)) = x_1
   POL(a(x_1)) = 1 + x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = 1 + x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   c(R(x1)) -> b(a(R(x1)))
   b(a(a(x1))) -> a(b(c(x1)))


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

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

   C(b(x1)) -> B(a(x1))

The TRS R consists of the following rules:

   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   c(R(x1)) -> b(a(R(x1)))
   b(a(a(x1))) -> a(b(c(x1)))

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

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

(11)
TRUE

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

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

   L^1(a(a(x1))) -> L^1(a(b(c(x1))))

The TRS R consists of the following rules:

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   L(a(a(x1))) -> L(a(b(c(x1))))
   c(R(x1)) -> b(a(R(x1)))

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

(13) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

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

   L^1(a(a(x1))) -> L^1(a(b(c(x1))))

The TRS R consists of the following rules:

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   L(a(a(x1))) -> L(a(b(c(x1))))
   c(R(x1)) -> b(a(R(x1)))

The set Q consists of the following terms:

   b(a(a(x0)))
   c(a(x0))
   c(b(x0))
   L(a(a(x0)))
   c(R(x0))

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

(15) 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.
----------------------------------------

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

   L^1(a(a(x1))) -> L^1(a(b(c(x1))))

The TRS R consists of the following rules:

   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   c(R(x1)) -> b(a(R(x1)))
   b(a(a(x1))) -> a(b(c(x1)))

The set Q consists of the following terms:

   b(a(a(x0)))
   c(a(x0))
   c(b(x0))
   L(a(a(x0)))
   c(R(x0))

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

(17) 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].

   L(a(a(x0)))


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

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

   L^1(a(a(x1))) -> L^1(a(b(c(x1))))

The TRS R consists of the following rules:

   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   c(R(x1)) -> b(a(R(x1)))
   b(a(a(x1))) -> a(b(c(x1)))

The set Q consists of the following terms:

   b(a(a(x0)))
   c(a(x0))
   c(b(x0))
   c(R(x0))

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

(19) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   L^1(a(a(x1))) -> L^1(a(b(c(x1))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( L^1_1(x_1) ) = 2x_1 + 2
POL( a_1(x_1) ) = x_1 + 1
POL( b_1(x_1) ) = max{0, x_1 - 1}
POL( c_1(x_1) ) = x_1 + 1
POL( R_1(x_1) ) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   c(R(x1)) -> b(a(R(x1)))
   b(a(a(x1))) -> a(b(c(x1)))


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

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

   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(a(x1))
   c(R(x1)) -> b(a(R(x1)))
   b(a(a(x1))) -> a(b(c(x1)))

The set Q consists of the following terms:

   b(a(a(x0)))
   c(a(x0))
   c(b(x0))
   c(R(x0))

We have to consider all minimal (P,Q,R)-chains.
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(21) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(22)
YES