/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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 w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 7 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES
    (28) QDP
        (29) UsableRulesProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) QReductionProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) TransformationProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) TransformationProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) TransformationProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) UsableRulesProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) QReductionProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) Induction-Processor [SOUND, 3 ms]
        (44) AND
            (45) QDP
                (46) PisEmptyProof [EQUIVALENT, 0 ms]
                (47) YES
            (48) QTRS
                (49) QTRSRRRProof [EQUIVALENT, 72 ms]
                (50) QTRS
                (51) QTRSRRRProof [EQUIVALENT, 0 ms]
                (52) QTRS
                (53) RisEmptyProof [EQUIVALENT, 0 ms]
                (54) YES


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

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

   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   del(x, nil) -> nil
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   first(nil) -> 0
   first(cons(x, xs)) -> x
   doublelist(nil) -> nil
   doublelist(cons(x, xs)) -> cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

Q is empty.

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

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

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

   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   del(x, nil) -> nil
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   first(nil) -> 0
   first(cons(x, xs)) -> x
   doublelist(nil) -> nil
   doublelist(cons(x, xs)) -> cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

   double(0)
   double(s(x0))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   first(nil)
   first(cons(x0, x1))
   doublelist(nil)
   doublelist(cons(x0, x1))


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

(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:

   DOUBLE(s(x)) -> DOUBLE(x)
   DEL(x, cons(y, xs)) -> IF(eq(x, y), x, y, xs)
   DEL(x, cons(y, xs)) -> EQ(x, y)
   IF(false, x, y, xs) -> DEL(x, xs)
   EQ(s(x), s(y)) -> EQ(x, y)
   DOUBLELIST(cons(x, xs)) -> DOUBLE(x)
   DOUBLELIST(cons(x, xs)) -> DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))
   DOUBLELIST(cons(x, xs)) -> DEL(first(cons(x, xs)), cons(x, xs))
   DOUBLELIST(cons(x, xs)) -> FIRST(cons(x, xs))

The TRS R consists of the following rules:

   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   del(x, nil) -> nil
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   first(nil) -> 0
   first(cons(x, xs)) -> x
   doublelist(nil) -> nil
   doublelist(cons(x, xs)) -> cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

   double(0)
   double(s(x0))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   first(nil)
   first(cons(x0, x1))
   doublelist(nil)
   doublelist(cons(x0, x1))

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:

   EQ(s(x), s(y)) -> EQ(x, y)

The TRS R consists of the following rules:

   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   del(x, nil) -> nil
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   first(nil) -> 0
   first(cons(x, xs)) -> x
   doublelist(nil) -> nil
   doublelist(cons(x, xs)) -> cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

   double(0)
   double(s(x0))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   first(nil)
   first(cons(x0, x1))
   doublelist(nil)
   doublelist(cons(x0, x1))

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

(8) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   EQ(s(x), s(y)) -> EQ(x, y)

R is empty.
The set Q consists of the following terms:

   double(0)
   double(s(x0))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   first(nil)
   first(cons(x0, x1))
   doublelist(nil)
   doublelist(cons(x0, x1))

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

(10) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   double(0)
   double(s(x0))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   first(nil)
   first(cons(x0, x1))
   doublelist(nil)
   doublelist(cons(x0, x1))


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

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

   EQ(s(x), s(y)) -> EQ(x, y)

R is empty.
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