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


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 81 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 2 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 23 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
    (19) QDP
        (20) UsableRulesProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (23) YES
    (24) QDP
        (25) UsableRulesProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (28) YES
    (29) QDP
        (30) UsableRulesProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (33) YES


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

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

   q0(0(x1)) -> 0'(q1(x1))
   q1(0(x1)) -> 0(q1(x1))
   q1(1'(x1)) -> 1'(q1(x1))
   0(q1(1(x1))) -> q2(0(1'(x1)))
   0'(q1(1(x1))) -> q2(0'(1'(x1)))
   1'(q1(1(x1))) -> q2(1'(1'(x1)))
   0(q2(0(x1))) -> q2(0(0(x1)))
   0'(q2(0(x1))) -> q2(0'(0(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   0'(q2(1'(x1))) -> q2(0'(1'(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))
   q2(0'(x1)) -> 0'(q0(x1))
   q0(1'(x1)) -> 1'(q3(x1))
   q3(1'(x1)) -> 1'(q3(x1))
   q3(b(x1)) -> b(q4(x1))

Q is empty.

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

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

   POL(0(x_1)) = x_1
   POL(0'(x_1)) = x_1
   POL(1(x_1)) = 3 + x_1
   POL(1'(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(q0(x_1)) = 2 + x_1
   POL(q1(x_1)) = x_1
   POL(q2(x_1)) = 3 + x_1
   POL(q3(x_1)) = 1 + x_1
   POL(q4(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   q0(0(x1)) -> 0'(q1(x1))
   q2(0'(x1)) -> 0'(q0(x1))
   q0(1'(x1)) -> 1'(q3(x1))
   q3(b(x1)) -> b(q4(x1))




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

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

   q1(0(x1)) -> 0(q1(x1))
   q1(1'(x1)) -> 1'(q1(x1))
   0(q1(1(x1))) -> q2(0(1'(x1)))
   0'(q1(1(x1))) -> q2(0'(1'(x1)))
   1'(q1(1(x1))) -> q2(1'(1'(x1)))
   0(q2(0(x1))) -> q2(0(0(x1)))
   0'(q2(0(x1))) -> q2(0'(0(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   0'(q2(1'(x1))) -> q2(0'(1'(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))
   q3(1'(x1)) -> 1'(q3(x1))

Q is empty.

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

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

   POL(0(x_1)) = x_1
   POL(0'(x_1)) = x_1
   POL(1(x_1)) = 1 + x_1
   POL(1'(x_1)) = x_1
   POL(q1(x_1)) = x_1
   POL(q2(x_1)) = x_1
   POL(q3(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   0(q1(1(x1))) -> q2(0(1'(x1)))
   0'(q1(1(x1))) -> q2(0'(1'(x1)))
   1'(q1(1(x1))) -> q2(1'(1'(x1)))




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

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

   q1(0(x1)) -> 0(q1(x1))
   q1(1'(x1)) -> 1'(q1(x1))
   0(q2(0(x1))) -> q2(0(0(x1)))
   0'(q2(0(x1))) -> q2(0'(0(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   0'(q2(1'(x1))) -> q2(0'(1'(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))
   q3(1'(x1)) -> 1'(q3(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:

   Q1(0(x1)) -> 0^1(q1(x1))
   Q1(0(x1)) -> Q1(x1)
   Q1(1'(x1)) -> 1'^1(q1(x1))
   Q1(1'(x1)) -> Q1(x1)
   0^1(q2(0(x1))) -> 0^1(0(x1))
   0'^1(q2(0(x1))) -> 0'^1(0(x1))
   1'^1(q2(0(x1))) -> 1'^1(0(x1))
   0^1(q2(1'(x1))) -> 0^1(1'(x1))
   0'^1(q2(1'(x1))) -> 0'^1(1'(x1))
   1'^1(q2(1'(x1))) -> 1'^1(1'(x1))
   Q3(1'(x1)) -> 1'^1(q3(x1))
   Q3(1'(x1)) -> Q3(x1)

The TRS R consists of the following rules:

   q1(0(x1)) -> 0(q1(x1))
   q1(1'(x1)) -> 1'(q1(x1))
   0(q2(0(x1))) -> q2(0(0(x1)))
   0'(q2(0(x1))) -> q2(0'(0(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   0'(q2(1'(x1))) -> q2(0'(1'(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))
   q3(1'(x1)) -> 1'(q3(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 5 SCCs with 3 less nodes.
----------------------------------------

(8)
Complex Obligation (AND)

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

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

   1'^1(q2(1'(x1))) -> 1'^1(1'(x1))
   1'^1(q2(0(x1))) -> 1'^1(0(x1))

The TRS R consists of the following rules:

   q1(0(x1)) -> 0(q1(x1))
   q1(1'(x1)) -> 1'(q1(x1))
   0(q2(0(x1))) -> q2(0(0(x1)))
   0'(q2(0(x1))) -> q2(0'(0(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   0'(q2(1'(x1))) -> q2(0'(1'(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))
   q3(1'(x1)) -> 1'(q3(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:

   1'^1(q2(1'(x1))) -> 1'^1(1'(x1))
   1'^1(q2(0(x1))) -> 1'^1(0(x1))

The TRS R consists of the following rules:

   0(q2(0(x1))) -> q2(0(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))

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:
*1'^1(q2(1'(x1))) -> 1'^1(1'(x1))
The graph contains the following edges 1 > 1


*1'^1(q2(0(x1))) -> 1'^1(0(x1))
The graph contains the following edges 1 > 1


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

(13)
YES

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

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

   Q3(1'(x1)) -> Q3(x1)

The TRS R consists of the following rules:

   q1(0(x1)) -> 0(q1(x1))
   q1(1'(x1)) -> 1'(q1(x1))
   0(q2(0(x1))) -> q2(0(0(x1)))
   0'(q2(0(x1))) -> q2(0'(0(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   0'(q2(1'(x1))) -> q2(0'(1'(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))
   q3(1'(x1)) -> 1'(q3(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:

   Q3(1'(x1)) -> Q3(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:
*Q3(1'(x1)) -> Q3(x1)
The graph contains the following edges 1 > 1


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

(18)
YES

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

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

   0'^1(q2(1'(x1))) -> 0'^1(1'(x1))
   0'^1(q2(0(x1))) -> 0'^1(0(x1))

The TRS R consists of the following rules:

   q1(0(x1)) -> 0(q1(x1))
   q1(1'(x1)) -> 1'(q1(x1))
   0(q2(0(x1))) -> q2(0(0(x1)))
   0'(q2(0(x1))) -> q2(0'(0(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   0'(q2(1'(x1))) -> q2(0'(1'(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))
   q3(1'(x1)) -> 1'(q3(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:

   0'^1(q2(1'(x1))) -> 0'^1(1'(x1))
   0'^1(q2(0(x1))) -> 0'^1(0(x1))

The TRS R consists of the following rules:

   0(q2(0(x1))) -> q2(0(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))

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:
*0'^1(q2(1'(x1))) -> 0'^1(1'(x1))
The graph contains the following edges 1 > 1


*0'^1(q2(0(x1))) -> 0'^1(0(x1))
The graph contains the following edges 1 > 1


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

(23)
YES

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

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

   0^1(q2(1'(x1))) -> 0^1(1'(x1))
   0^1(q2(0(x1))) -> 0^1(0(x1))

The TRS R consists of the following rules:

   q1(0(x1)) -> 0(q1(x1))
   q1(1'(x1)) -> 1'(q1(x1))
   0(q2(0(x1))) -> q2(0(0(x1)))
   0'(q2(0(x1))) -> q2(0'(0(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   0'(q2(1'(x1))) -> q2(0'(1'(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))
   q3(1'(x1)) -> 1'(q3(x1))

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

(25) 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.
----------------------------------------

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

   0^1(q2(1'(x1))) -> 0^1(1'(x1))
   0^1(q2(0(x1))) -> 0^1(0(x1))

The TRS R consists of the following rules:

   0(q2(0(x1))) -> q2(0(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))

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

(27) 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:
*0^1(q2(1'(x1))) -> 0^1(1'(x1))
The graph contains the following edges 1 > 1


*0^1(q2(0(x1))) -> 0^1(0(x1))
The graph contains the following edges 1 > 1


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

(28)
YES

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

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   Q1(1'(x1)) -> Q1(x1)
   Q1(0(x1)) -> Q1(x1)

The TRS R consists of the following rules:

   q1(0(x1)) -> 0(q1(x1))
   q1(1'(x1)) -> 1'(q1(x1))
   0(q2(0(x1))) -> q2(0(0(x1)))
   0'(q2(0(x1))) -> q2(0'(0(x1)))
   1'(q2(0(x1))) -> q2(1'(0(x1)))
   0(q2(1'(x1))) -> q2(0(1'(x1)))
   0'(q2(1'(x1))) -> q2(0'(1'(x1)))
   1'(q2(1'(x1))) -> q2(1'(1'(x1)))
   q3(1'(x1)) -> 1'(q3(x1))

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

(30) 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.
----------------------------------------

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   Q1(1'(x1)) -> Q1(x1)
   Q1(0(x1)) -> Q1(x1)

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

(32) 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:
*Q1(1'(x1)) -> Q1(x1)
The graph contains the following edges 1 > 1


*Q1(0(x1)) -> Q1(x1)
The graph contains the following edges 1 > 1


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

(33)
YES