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


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 1 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) TransformationProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPOrderProof [EQUIVALENT, 189 ms]
        (24) QDP
        (25) QDPOrderProof [EQUIVALENT, 105 ms]
        (26) QDP
        (27) QDPOrderProof [EQUIVALENT, 25 ms]
        (28) QDP
        (29) QDPOrderProof [EQUIVALENT, 2404 ms]
        (30) QDP
        (31) PisEmptyProof [EQUIVALENT, 0 ms]
        (32) YES


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

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

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

Q is empty.

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

(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:

   .^1(.(x, y), z) -> .^1(x, .(y, z))
   .^1(.(x, y), z) -> .^1(y, z)
   A(f(x)) -> F(a(x))
   A(f(x)) -> A(x)
   A(.(x, y)) -> .^1(a(x), y)
   A(.(x, y)) -> A(x)
   A(b1(x)) -> B1(a(x))
   A(b1(x)) -> A(x)
   F(b(x)) -> F(x)
   .^1(b(x), y) -> .^1(x, y)
   A(f(.(0, x))) -> B1(.(f(.(0, x)), .(0, f(x))))
   A(f(.(0, x))) -> .^1(f(.(0, x)), .(0, f(x)))
   A(f(.(0, x))) -> .^1(0, f(x))
   A(f(.(0, x))) -> F(x)
   A(f(0)) -> B1(.(f(0), 0))
   A(f(0)) -> .^1(f(0), 0)
   F(.(0, x)) -> .^1(0, f(x))
   F(.(0, x)) -> F(x)
   C(b(x)) -> C(a(x))
   C(b(x)) -> A(x)
   A(b(x)) -> A(x)

The TRS R consists of the following rules:

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

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 4 SCCs with 11 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   .^1(.(x, y), z) -> .^1(y, z)
   .^1(.(x, y), z) -> .^1(x, .(y, z))
   .^1(b(x), y) -> .^1(x, y)

The TRS R consists of the following rules:

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

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:

   .^1(.(x, y), z) -> .^1(y, z)
   .^1(.(x, y), z) -> .^1(x, .(y, z))
   .^1(b(x), y) -> .^1(x, y)

The TRS R consists of the following rules:

   .(.(x, y), z) -> .(x, .(y, z))
   .(b(x), y) -> b(.(x, y))

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

(8) 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:
*.^1(.(x, y), z) -> .^1(y, z)
The graph contains the following edges 1 > 1, 2 >= 2


*.^1(.(x, y), z) -> .^1(x, .(y, z))
The graph contains the following edges 1 > 1


*.^1(b(x), y) -> .^1(x, y)
The graph contains the following edges 1 > 1, 2 >= 2


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

(9)
YES

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

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(.(0, x)) -> F(x)
   F(b(x)) -> F(x)

The TRS R consists of the following rules:

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

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

(11) 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.
----------------------------------------

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

   F(.(0, x)) -> F(x)
   F(b(x)) -> F(x)

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

(13) 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(.(0, x)) -> F(x)
The graph contains the following edges 1 > 1


*F(b(x)) -> F(x)
The graph contains the following edges 1 > 1


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

(14)
YES

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

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

   A(.(x, y)) -> A(x)
   A(f(x)) -> A(x)
   A(b1(x)) -> A(x)
   A(b(x)) -> A(x)

The TRS R consists of the following rules:

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

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

(16) 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.
----------------------------------------

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

   A(.(x, y)) -> A(x)
   A(f(x)) -> A(x)
   A(b1(x)) -> A(x)
   A(b(x)) -> A(x)

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

(18) 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:
*A(.(x, y)) -> A(x)
The graph contains the following edges 1 > 1


*A(f(x)) -> A(x)
The graph contains the following edges 1 > 1


*A(b1(x)) -> A(x)
The graph contains the following edges 1 > 1


*A(b(x)) -> A(x)
The graph contains the following edges 1 > 1


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

(19)
YES

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

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

   C(b(x)) -> C(a(x))

The TRS R consists of the following rules:

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

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

(21) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule C(b(x)) -> C(a(x)) at position [0] we obtained the following new rules [LPAR04]:

   (C(b(f(x0))) -> C(f(a(x0))),C(b(f(x0))) -> C(f(a(x0))))
   (C(b(.(x0, x1))) -> C(.(a(x0), x1)),C(b(.(x0, x1))) -> C(.(a(x0), x1)))
   (C(b(b1(x0))) -> C(b1(a(x0))),C(b(b1(x0))) -> C(b1(a(x0))))
   (C(b(f(.(0, x0)))) -> C(b1(.(f(.(0, x0)), .(0, f(x0))))),C(b(f(.(0, x0)))) -> C(b1(.(f(.(0, x0)), .(0, f(x0))))))
   (C(b(f(0))) -> C(b1(.(f(0), 0))),C(b(f(0))) -> C(b1(.(f(0), 0))))
   (C(b(b(x0))) -> C(b(a(x0))),C(b(b(x0))) -> C(b(a(x0))))


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

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

   C(b(f(x0))) -> C(f(a(x0)))
   C(b(.(x0, x1))) -> C(.(a(x0), x1))
   C(b(b1(x0))) -> C(b1(a(x0)))
   C(b(f(.(0, x0)))) -> C(b1(.(f(.(0, x0)), .(0, f(x0)))))
   C(b(f(0))) -> C(b1(.(f(0), 0)))
   C(b(b(x0))) -> C(b(a(x0)))

The TRS R consists of the following rules:

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

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

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


The following pairs can be oriented strictly and are deleted.

   C(b(f(0))) -> C(b1(.(f(0), 0)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
C(x1)  =  C(x1)

b(x1)  =  x1

f(x1)  =  x1

a(x1)  =  x1

.(x1, x2)  =  .

b1(x1)  =  x1

0  =  0


Recursive path order with status [RPO].
Quasi-Precedence: C_1 > 0 > .

Status: C_1: [1]
.: []
0: multiset status


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

   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   a(b(x)) -> b(a(x))
   f(b(x)) -> b(f(x))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   .(.(x, y), z) -> .(x, .(y, z))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))


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

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

   C(b(f(x0))) -> C(f(a(x0)))
   C(b(.(x0, x1))) -> C(.(a(x0), x1))
   C(b(b1(x0))) -> C(b1(a(x0)))
   C(b(f(.(0, x0)))) -> C(b1(.(f(.(0, x0)), .(0, f(x0)))))
   C(b(b(x0))) -> C(b(a(x0)))

The TRS R consists of the following rules:

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

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

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


The following pairs can be oriented strictly and are deleted.

   C(b(f(.(0, x0)))) -> C(b1(.(f(.(0, x0)), .(0, f(x0)))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( ._2(x_1, x_2) ) = 0
POL( C_1(x_1) ) = 2x_1
POL( b_1(x_1) ) = x_1
POL( b1_1(x_1) ) = x_1
POL( f_1(x_1) ) = 2
POL( a_1(x_1) ) = x_1
POL( 0 ) = 0

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

   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   a(b(x)) -> b(a(x))
   f(b(x)) -> b(f(x))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   .(.(x, y), z) -> .(x, .(y, z))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))


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

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

   C(b(f(x0))) -> C(f(a(x0)))
   C(b(.(x0, x1))) -> C(.(a(x0), x1))
   C(b(b1(x0))) -> C(b1(a(x0)))
   C(b(b(x0))) -> C(b(a(x0)))

The TRS R consists of the following rules:

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

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

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


The following pairs can be oriented strictly and are deleted.

   C(b(b1(x0))) -> C(b1(a(x0)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

   <<<
 POL(C(x_1)) =  	[[0]] 	 +  	[[1, 0]] 	* 	x_1
>>>

   <<<
 POL(b(x_1)) =  	[[0], [0]] 	 +  	[[1, 1], [0, 0]] 	* 	x_1
>>>

   <<<
 POL(f(x_1)) =  	[[1], [0]] 	 +  	[[1, 0], [0, 1]] 	* 	x_1
>>>

   <<<
 POL(a(x_1)) =  	[[0], [0]] 	 +  	[[1, 0], [0, 1]] 	* 	x_1
>>>

   <<<
 POL(.(x_1, x_2)) =  	[[0], [0]] 	 +  	[[1, 0], [0, 1]] 	* 	x_1 	 +  	[[0, 0], [0, 0]] 	* 	x_2
>>>

   <<<
 POL(b1(x_1)) =  	[[0], [1]] 	 +  	[[1, 0], [0, 0]] 	* 	x_1
>>>

   <<<
 POL(0) =  	[[0], [1]]
>>>


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

   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   a(b(x)) -> b(a(x))
   f(b(x)) -> b(f(x))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   .(.(x, y), z) -> .(x, .(y, z))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))


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

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

   C(b(f(x0))) -> C(f(a(x0)))
   C(b(.(x0, x1))) -> C(.(a(x0), x1))
   C(b(b(x0))) -> C(b(a(x0)))

The TRS R consists of the following rules:

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

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

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


The following pairs can be oriented strictly and are deleted.

   C(b(f(x0))) -> C(f(a(x0)))
   C(b(.(x0, x1))) -> C(.(a(x0), x1))
   C(b(b(x0))) -> C(b(a(x0)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

   <<<
 POL(C(x_1)) =  	[[0]] 	 +  	[[0, 1]] 	* 	x_1
>>>

   <<<
 POL(b(x_1)) =  	[[0], [1]] 	 +  	[[0, 1], [0, 2]] 	* 	x_1
>>>

   <<<
 POL(f(x_1)) =  	[[0], [0]] 	 +  	[[1, 1], [0, 3]] 	* 	x_1
>>>

   <<<
 POL(a(x_1)) =  	[[1], [0]] 	 +  	[[2, 0], [0, 2]] 	* 	x_1
>>>

   <<<
 POL(.(x_1, x_2)) =  	[[0], [0]] 	 +  	[[1, 0], [0, 1]] 	* 	x_1 	 +  	[[0, 0], [0, 0]] 	* 	x_2
>>>

   <<<
 POL(b1(x_1)) =  	[[1], [2]] 	 +  	[[2, 0], [2, 1]] 	* 	x_1
>>>

   <<<
 POL(0) =  	[[0], [3]]
>>>


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

   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   a(b(x)) -> b(a(x))
   f(b(x)) -> b(f(x))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   .(.(x, y), z) -> .(x, .(y, z))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))


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

(30)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   .(.(x, y), z) -> .(x, .(y, z))
   a(f(x)) -> f(a(x))
   a(.(x, y)) -> .(a(x), y)
   a(b1(x)) -> b1(a(x))
   f(b(x)) -> b(f(x))
   .(b(x), y) -> b(.(x, y))
   b1(b(x)) -> b(b(x))
   a(f(.(0, x))) -> b1(.(f(.(0, x)), .(0, f(x))))
   a(f(0)) -> b1(.(f(0), 0))
   f(.(0, x)) -> b(.(0, f(x)))
   f(0) -> b(0)
   c(b(x)) -> c(a(x))
   a(b(x)) -> b(a(x))

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

(31) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(32)
YES