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


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


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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 19 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPOrderProof [EQUIVALENT, 19 ms]
        (11) QDP
        (12) PisEmptyProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) UsableRulesProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (23) YES
    (24) QDP
        (25) QDPOrderProof [EQUIVALENT, 445 ms]
        (26) QDP
        (27) QDPOrderProof [EQUIVALENT, 11 ms]
        (28) QDP
        (29) PisEmptyProof [EQUIVALENT, 0 ms]
        (30) YES


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

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

   r1(a(x1)) -> a(a(a(r1(x1))))
   r2(a(x1)) -> a(a(a(r2(x1))))
   a(l1(x1)) -> l1(a(a(a(x1))))
   a(a(l2(x1))) -> l2(a(a(x1)))
   r1(b(x1)) -> l1(b(x1))
   r2(b(x1)) -> l2(a(b(x1)))
   b(l1(x1)) -> b(r2(x1))
   b(l2(x1)) -> b(r1(x1))
   a(a(x1)) -> 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:

   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   l1(a(x1)) -> a(a(a(l1(x1))))
   l2(a(a(x1))) -> a(a(l2(x1)))
   b(r1(x1)) -> b(l1(x1))
   b(r2(x1)) -> b(a(l2(x1)))
   l1(b(x1)) -> r2(b(x1))
   l2(b(x1)) -> r1(b(x1))
   a(a(x1)) -> x1

Q is empty.

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

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(r1(x1)) -> A(a(a(x1)))
   A(r1(x1)) -> A(a(x1))
   A(r1(x1)) -> A(x1)
   A(r2(x1)) -> A(a(a(x1)))
   A(r2(x1)) -> A(a(x1))
   A(r2(x1)) -> A(x1)
   L1(a(x1)) -> A(a(a(l1(x1))))
   L1(a(x1)) -> A(a(l1(x1)))
   L1(a(x1)) -> A(l1(x1))
   L1(a(x1)) -> L1(x1)
   L2(a(a(x1))) -> A(a(l2(x1)))
   L2(a(a(x1))) -> A(l2(x1))
   L2(a(a(x1))) -> L2(x1)
   B(r1(x1)) -> B(l1(x1))
   B(r1(x1)) -> L1(x1)
   B(r2(x1)) -> B(a(l2(x1)))
   B(r2(x1)) -> A(l2(x1))
   B(r2(x1)) -> L2(x1)

The TRS R consists of the following rules:

   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   l1(a(x1)) -> a(a(a(l1(x1))))
   l2(a(a(x1))) -> a(a(l2(x1)))
   b(r1(x1)) -> b(l1(x1))
   b(r2(x1)) -> b(a(l2(x1)))
   l1(b(x1)) -> r2(b(x1))
   l2(b(x1)) -> r1(b(x1))
   a(a(x1)) -> x1

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 8 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   A(r1(x1)) -> A(a(x1))
   A(r1(x1)) -> A(a(a(x1)))
   A(r1(x1)) -> A(x1)
   A(r2(x1)) -> A(a(a(x1)))
   A(r2(x1)) -> A(a(x1))
   A(r2(x1)) -> A(x1)

The TRS R consists of the following rules:

   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   l1(a(x1)) -> a(a(a(l1(x1))))
   l2(a(a(x1))) -> a(a(l2(x1)))
   b(r1(x1)) -> b(l1(x1))
   b(r2(x1)) -> b(a(l2(x1)))
   l1(b(x1)) -> r2(b(x1))
   l2(b(x1)) -> r1(b(x1))
   a(a(x1)) -> x1

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

(8) 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.
----------------------------------------

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

   A(r1(x1)) -> A(a(x1))
   A(r1(x1)) -> A(a(a(x1)))
   A(r1(x1)) -> A(x1)
   A(r2(x1)) -> A(a(a(x1)))
   A(r2(x1)) -> A(a(x1))
   A(r2(x1)) -> A(x1)

The TRS R consists of the following rules:

   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   a(a(x1)) -> x1

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

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


The following pairs can be oriented strictly and are deleted.

   A(r1(x1)) -> A(a(x1))
   A(r1(x1)) -> A(a(a(x1)))
   A(r1(x1)) -> A(x1)
   A(r2(x1)) -> A(a(a(x1)))
   A(r2(x1)) -> A(a(x1))
   A(r2(x1)) -> A(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( A_1(x_1) ) = x_1 + 1
POL( a_1(x_1) ) = x_1
POL( r1_1(x_1) ) = 2x_1 + 1
POL( r2_1(x_1) ) = 2x_1 + 1

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

   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   a(a(x1)) -> x1


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

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

   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   a(a(x1)) -> x1

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

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

(13)
YES

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

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

   L2(a(a(x1))) -> L2(x1)

The TRS R consists of the following rules:

   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   l1(a(x1)) -> a(a(a(l1(x1))))
   l2(a(a(x1))) -> a(a(l2(x1)))
   b(r1(x1)) -> b(l1(x1))
   b(r2(x1)) -> b(a(l2(x1)))
   l1(b(x1)) -> r2(b(x1))
   l2(b(x1)) -> r1(b(x1))
   a(a(x1)) -> 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:

   L2(a(a(x1))) -> L2(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:
*L2(a(a(x1))) -> L2(x1)
The graph contains the following edges 1 > 1


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

(18)
YES

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

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   L1(a(x1)) -> L1(x1)

The TRS R consists of the following rules:

   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   l1(a(x1)) -> a(a(a(l1(x1))))
   l2(a(a(x1))) -> a(a(l2(x1)))
   b(r1(x1)) -> b(l1(x1))
   b(r2(x1)) -> b(a(l2(x1)))
   l1(b(x1)) -> r2(b(x1))
   l2(b(x1)) -> r1(b(x1))
   a(a(x1)) -> x1

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

(20) 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.
----------------------------------------

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   L1(a(x1)) -> L1(x1)

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

(22) 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:
*L1(a(x1)) -> L1(x1)
The graph contains the following edges 1 > 1


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

(23)
YES

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

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

   B(r2(x1)) -> B(a(l2(x1)))
   B(r1(x1)) -> B(l1(x1))

The TRS R consists of the following rules:

   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   l1(a(x1)) -> a(a(a(l1(x1))))
   l2(a(a(x1))) -> a(a(l2(x1)))
   b(r1(x1)) -> b(l1(x1))
   b(r2(x1)) -> b(a(l2(x1)))
   l1(b(x1)) -> r2(b(x1))
   l2(b(x1)) -> r1(b(x1))
   a(a(x1)) -> x1

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.

   B(r2(x1)) -> B(a(l2(x1)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(B(x_1)) =  	[[0A]] 	 +  	[[-I, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(r2(x_1)) =  	[[1A], [1A], [0A]] 	 +  	[[1A, 0A, 0A], [0A, 1A, 0A], [0A, -I, 1A]] 	* 	x_1
>>>

   <<<
 POL(a(x_1)) =  	[[0A], [-I], [-I]] 	 +  	[[-I, 0A, 0A], [0A, -I, -I], [0A, -I, -I]] 	* 	x_1
>>>

   <<<
 POL(l2(x_1)) =  	[[0A], [0A], [0A]] 	 +  	[[-I, 0A, 0A], [0A, -I, 1A], [0A, 1A, 0A]] 	* 	x_1
>>>

   <<<
 POL(r1(x_1)) =  	[[1A], [0A], [1A]] 	 +  	[[1A, 0A, 0A], [0A, -I, 1A], [0A, 1A, 0A]] 	* 	x_1
>>>

   <<<
 POL(l1(x_1)) =  	[[0A], [0A], [0A]] 	 +  	[[1A, -I, 0A], [0A, 1A, 0A], [0A, -I, 1A]] 	* 	x_1
>>>

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


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

   l2(a(a(x1))) -> a(a(l2(x1)))
   l2(b(x1)) -> r1(b(x1))
   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   a(a(x1)) -> x1
   l1(a(x1)) -> a(a(a(l1(x1))))
   l1(b(x1)) -> r2(b(x1))
   b(r2(x1)) -> b(a(l2(x1)))
   b(r1(x1)) -> b(l1(x1))


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

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

   B(r1(x1)) -> B(l1(x1))

The TRS R consists of the following rules:

   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   l1(a(x1)) -> a(a(a(l1(x1))))
   l2(a(a(x1))) -> a(a(l2(x1)))
   b(r1(x1)) -> b(l1(x1))
   b(r2(x1)) -> b(a(l2(x1)))
   l1(b(x1)) -> r2(b(x1))
   l2(b(x1)) -> r1(b(x1))
   a(a(x1)) -> x1

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.

   B(r1(x1)) -> B(l1(x1))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(B(x_1)) = x_1
   POL(a(x_1)) = x_1
   POL(b(x_1)) = 1
   POL(l1(x_1)) = 0
   POL(l2(x_1)) = 1 + x_1
   POL(r1(x_1)) = 1
   POL(r2(x_1)) = 0

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

   l1(a(x1)) -> a(a(a(l1(x1))))
   l1(b(x1)) -> r2(b(x1))
   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   a(a(x1)) -> x1


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

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

   a(r1(x1)) -> r1(a(a(a(x1))))
   a(r2(x1)) -> r2(a(a(a(x1))))
   l1(a(x1)) -> a(a(a(l1(x1))))
   l2(a(a(x1))) -> a(a(l2(x1)))
   b(r1(x1)) -> b(l1(x1))
   b(r2(x1)) -> b(a(l2(x1)))
   l1(b(x1)) -> r2(b(x1))
   l2(b(x1)) -> r1(b(x1))
   a(a(x1)) -> x1

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

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

(30)
YES