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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 47 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) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (16) YES
    (17) QDP
        (18) UsableRulesProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (21) YES
    (22) QDP
        (23) UsableRulesProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) MRRProof [EQUIVALENT, 7 ms]
        (26) QDP
        (27) QDPOrderProof [EQUIVALENT, 35 ms]
        (28) QDP
        (29) PisEmptyProof [EQUIVALENT, 0 ms]
        (30) YES


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

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

   t(f(x1)) -> t(c(n(x1)))
   n(f(x1)) -> f(n(x1))
   o(f(x1)) -> f(o(x1))
   n(s(x1)) -> f(s(x1))
   o(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(x1)

Q is empty.

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

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

   POL(c(x_1)) = x_1
   POL(f(x_1)) = x_1
   POL(n(x_1)) = x_1
   POL(o(x_1)) = 1 + x_1
   POL(s(x_1)) = x_1
   POL(t(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   o(s(x1)) -> f(s(x1))




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

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

   t(f(x1)) -> t(c(n(x1)))
   n(f(x1)) -> f(n(x1))
   o(f(x1)) -> f(o(x1))
   n(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(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:

   T(f(x1)) -> T(c(n(x1)))
   T(f(x1)) -> C(n(x1))
   T(f(x1)) -> N(x1)
   N(f(x1)) -> N(x1)
   O(f(x1)) -> O(x1)
   C(f(x1)) -> C(x1)
   C(n(x1)) -> N(c(x1))
   C(n(x1)) -> C(x1)
   C(o(x1)) -> O(c(x1))
   C(o(x1)) -> C(x1)

The TRS R consists of the following rules:

   t(f(x1)) -> t(c(n(x1)))
   n(f(x1)) -> f(n(x1))
   o(f(x1)) -> f(o(x1))
   n(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(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 4 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   O(f(x1)) -> O(x1)

The TRS R consists of the following rules:

   t(f(x1)) -> t(c(n(x1)))
   n(f(x1)) -> f(n(x1))
   o(f(x1)) -> f(o(x1))
   n(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(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:

   O(f(x1)) -> O(x1)

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

(10) 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:
*O(f(x1)) -> O(x1)
The graph contains the following edges 1 > 1


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

(11)
YES

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

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

   N(f(x1)) -> N(x1)

The TRS R consists of the following rules:

   t(f(x1)) -> t(c(n(x1)))
   n(f(x1)) -> f(n(x1))
   o(f(x1)) -> f(o(x1))
   n(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(x1)

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

(13) 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.
----------------------------------------

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

   N(f(x1)) -> N(x1)

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

(15) 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:
*N(f(x1)) -> N(x1)
The graph contains the following edges 1 > 1


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

(16)
YES

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

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

   C(n(x1)) -> C(x1)
   C(f(x1)) -> C(x1)
   C(o(x1)) -> C(x1)

The TRS R consists of the following rules:

   t(f(x1)) -> t(c(n(x1)))
   n(f(x1)) -> f(n(x1))
   o(f(x1)) -> f(o(x1))
   n(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(x1)

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

(18) 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.
----------------------------------------

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

   C(n(x1)) -> C(x1)
   C(f(x1)) -> C(x1)
   C(o(x1)) -> C(x1)

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

(20) 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:
*C(n(x1)) -> C(x1)
The graph contains the following edges 1 > 1


*C(f(x1)) -> C(x1)
The graph contains the following edges 1 > 1


*C(o(x1)) -> C(x1)
The graph contains the following edges 1 > 1


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

(21)
YES

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

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

   T(f(x1)) -> T(c(n(x1)))

The TRS R consists of the following rules:

   t(f(x1)) -> t(c(n(x1)))
   n(f(x1)) -> f(n(x1))
   o(f(x1)) -> f(o(x1))
   n(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(x1)

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

(23) 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.
----------------------------------------

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

   T(f(x1)) -> T(c(n(x1)))

The TRS R consists of the following rules:

   n(f(x1)) -> f(n(x1))
   n(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(x1)
   o(f(x1)) -> f(o(x1))

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

(25) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   o(f(x1)) -> f(o(x1))

Used ordering: Polynomial interpretation [POLO]:

   POL(T(x_1)) = 2*x_1
   POL(c(x_1)) = x_1
   POL(f(x_1)) = 1 + 2*x_1
   POL(n(x_1)) = 1 + 2*x_1
   POL(o(x_1)) = 1 + 3*x_1
   POL(s(x_1)) = x_1


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

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

   T(f(x1)) -> T(c(n(x1)))

The TRS R consists of the following rules:

   n(f(x1)) -> f(n(x1))
   n(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(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.

   T(f(x1)) -> T(c(n(x1)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

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

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

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

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

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

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


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

   n(f(x1)) -> f(n(x1))
   n(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(x1)


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

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

   n(f(x1)) -> f(n(x1))
   n(s(x1)) -> f(s(x1))
   c(f(x1)) -> f(c(x1))
   c(n(x1)) -> n(c(x1))
   c(o(x1)) -> o(c(x1))
   c(o(x1)) -> o(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