/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination of the given CSR could be proven:

(0) CSR
(1) CSRInnermostProof [EQUIVALENT, 0 ms]
(2) CSR
(3) CSDependencyPairsProof [EQUIVALENT, 0 ms]
(4) QCSDP
(5) QCSDependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QCSDP
        (8) QCSDPSubtermProof [EQUIVALENT, 21 ms]
        (9) QCSDP
        (10) PIsEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QCSDP
        (13) QCSDPSubtermProof [EQUIVALENT, 0 ms]
        (14) QCSDP
        (15) PIsEmptyProof [EQUIVALENT, 0 ms]
        (16) YES
    (17) QCSDP
        (18) QCSDPSubtermProof [EQUIVALENT, 0 ms]
        (19) QCSDP
        (20) PIsEmptyProof [EQUIVALENT, 0 ms]
        (21) YES


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

(0)
Obligation:
Context-sensitive rewrite system:
The TRS R consists of the following rules:

   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

The replacement map contains the following entries:

terms: {1}
cons: {1}
recip: {1}
sqr: {1}
s: {1}
0: empty set
add: {1, 2}
dbl: {1}
first: {1, 2}
nil: empty set

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

(1) CSRInnermostProof (EQUIVALENT)
The CSR is orthogonal. By [CS_Inn] we can switch to innermost.
----------------------------------------

(2)
Obligation:
Context-sensitive rewrite system:
The TRS R consists of the following rules:

   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

The replacement map contains the following entries:

terms: {1}
cons: {1}
recip: {1}
sqr: {1}
s: {1}
0: empty set
add: {1, 2}
dbl: {1}
first: {1, 2}
nil: empty set


Innermost Strategy.

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

(3) CSDependencyPairsProof (EQUIVALENT)
Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.
----------------------------------------

(4)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, SQR_1, TERMS_1, ADD_2, DBL_1} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.

The ordinary context-sensitive dependency pairs DP_o are:
   TERMS(N) -> SQR(N)
   SQR(s(X)) -> ADD(sqr(X), dbl(X))
   SQR(s(X)) -> SQR(X)
   SQR(s(X)) -> DBL(X)
   DBL(s(X)) -> DBL(X)
   ADD(s(X), Y) -> ADD(X, Y)

The TRS R consists of the following rules:

   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

The set Q consists of the following terms:

   terms(x0)
   sqr(0)
   sqr(s(x0))
   dbl(0)
   dbl(s(x0))
   add(0, x0)
   add(s(x0), x1)
   first(0, x0)
   first(s(x0), cons(x1, x2))


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

(5) QCSDependencyGraphProof (EQUIVALENT)
The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 3 SCCs with 3 less nodes.

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

(6)
Complex Obligation (AND)

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

(7)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, DBL_1} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.

The TRS P consists of the following rules:

   DBL(s(X)) -> DBL(X)

The TRS R consists of the following rules:

   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

The set Q consists of the following terms:

   terms(x0)
   sqr(0)
   sqr(s(x0))
   dbl(0)
   dbl(s(x0))
   add(0, x0)
   add(s(x0), x1)
   first(0, x0)
   first(s(x0), cons(x1, x2))


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

(8) QCSDPSubtermProof (EQUIVALENT)
We use the subterm processor [DA_EMMES].


The following pairs can be oriented strictly and are deleted.

   DBL(s(X)) -> DBL(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering:  Combined order from the following AFS and order.
DBL(x1)  =  x1


Subterm Order

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

(9)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

The set Q consists of the following terms:

   terms(x0)
   sqr(0)
   sqr(s(x0))
   dbl(0)
   dbl(s(x0))
   add(0, x0)
   add(s(x0), x1)
   first(0, x0)
   first(s(x0), cons(x1, x2))


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

(10) PIsEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
----------------------------------------

(11)
YES

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

(12)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, ADD_2} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.

The TRS P consists of the following rules:

   ADD(s(X), Y) -> ADD(X, Y)

The TRS R consists of the following rules:

   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

The set Q consists of the following terms:

   terms(x0)
   sqr(0)
   sqr(s(x0))
   dbl(0)
   dbl(s(x0))
   add(0, x0)
   add(s(x0), x1)
   first(0, x0)
   first(s(x0), cons(x1, x2))


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

(13) QCSDPSubtermProof (EQUIVALENT)
We use the subterm processor [DA_EMMES].


The following pairs can be oriented strictly and are deleted.

   ADD(s(X), Y) -> ADD(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering:  Combined order from the following AFS and order.
ADD(x1, x2)  =  x1


Subterm Order

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

(14)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

The set Q consists of the following terms:

   terms(x0)
   sqr(0)
   sqr(s(x0))
   dbl(0)
   dbl(s(x0))
   add(0, x0)
   add(s(x0), x1)
   first(0, x0)
   first(s(x0), cons(x1, x2))


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

(15) PIsEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
----------------------------------------

(16)
YES

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

(17)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2, SQR_1} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.

The TRS P consists of the following rules:

   SQR(s(X)) -> SQR(X)

The TRS R consists of the following rules:

   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

The set Q consists of the following terms:

   terms(x0)
   sqr(0)
   sqr(s(x0))
   dbl(0)
   dbl(s(x0))
   add(0, x0)
   add(s(x0), x1)
   first(0, x0)
   first(s(x0), cons(x1, x2))


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

(18) QCSDPSubtermProof (EQUIVALENT)
We use the subterm processor [DA_EMMES].


The following pairs can be oriented strictly and are deleted.

   SQR(s(X)) -> SQR(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering:  Combined order from the following AFS and order.
SQR(x1)  =  x1


Subterm Order

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

(19)
Obligation:
Q-restricted context-sensitive dependency pair problem:
The symbols in {terms_1, recip_1, sqr_1, s_1, add_2, dbl_1, first_2} are replacing on all positions.
For all symbols f in {cons_2} we have mu(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

   terms(N) -> cons(recip(sqr(N)), terms(s(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))

The set Q consists of the following terms:

   terms(x0)
   sqr(0)
   sqr(s(x0))
   dbl(0)
   dbl(s(x0))
   add(0, x0)
   add(s(x0), x1)
   first(0, x0)
   first(s(x0), cons(x1, x2))


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

(20) PIsEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
----------------------------------------

(21)
YES