YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) AAECC Innermost [EQUIVALENT, 27 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 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) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (34) YES
    (35) QDP
        (36) UsableRulesProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) QReductionProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) NonInfProof [EQUIVALENT, 108 ms]
        (41) AND
            (42) QDP
                (43) DependencyGraphProof [EQUIVALENT, 0 ms]
                (44) TRUE
            (45) QDP
                (46) DependencyGraphProof [EQUIVALENT, 0 ms]
                (47) TRUE
    (48) QDP
        (49) UsableRulesProof [EQUIVALENT, 0 ms]
        (50) QDP
        (51) QReductionProof [EQUIVALENT, 0 ms]
        (52) QDP
        (53) TransformationProof [EQUIVALENT, 0 ms]
        (54) QDP
        (55) DependencyGraphProof [EQUIVALENT, 0 ms]
        (56) QDP
        (57) UsableRulesProof [EQUIVALENT, 0 ms]
        (58) QDP
        (59) QReductionProof [EQUIVALENT, 0 ms]
        (60) QDP
        (61) QDPOrderProof [EQUIVALENT, 0 ms]
        (62) QDP
        (63) DependencyGraphProof [EQUIVALENT, 0 ms]
        (64) TRUE


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

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

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))
   maplog(xs) -> mapIter(xs, nil)
   mapIter(xs, ys) -> ifmap(isempty(xs), xs, ys)
   ifmap(true, xs, ys) -> ys
   ifmap(false, xs, ys) -> mapIter(droplast(xs), cons(log(last(xs)), ys))
   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   a -> b
   a -> c

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))
   ifmap(true, xs, ys) -> ys
   ifmap(false, xs, ys) -> mapIter(droplast(xs), cons(log(last(xs)), ys))
   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   maplog(xs) -> mapIter(xs, nil)
   mapIter(xs, ys) -> ifmap(isempty(xs), xs, ys)

The TRS R 2 is 
   a -> b
   a -> c

The signature Sigma is {a, b, c}
----------------------------------------

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

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))
   maplog(xs) -> mapIter(xs, nil)
   mapIter(xs, ys) -> ifmap(isempty(xs), xs, ys)
   ifmap(true, xs, ys) -> ys
   ifmap(false, xs, ys) -> mapIter(droplast(xs), cons(log(last(xs)), ys))
   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   a -> b
   a -> c

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a


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

(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:

   LE(s(x), s(y)) -> LE(x, y)
   DOUBLE(s(x)) -> DOUBLE(x)
   LOG(s(x)) -> LOOP(s(x), s(0), 0)
   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)
   LOOP(x, s(y), z) -> LE(x, s(y))
   IF(false, x, y, z) -> LOOP(x, double(y), s(z))
   IF(false, x, y, z) -> DOUBLE(y)
   MAPLOG(xs) -> MAPITER(xs, nil)
   MAPITER(xs, ys) -> IFMAP(isempty(xs), xs, ys)
   MAPITER(xs, ys) -> ISEMPTY(xs)
   IFMAP(false, xs, ys) -> MAPITER(droplast(xs), cons(log(last(xs)), ys))
   IFMAP(false, xs, ys) -> DROPLAST(xs)
   IFMAP(false, xs, ys) -> LOG(last(xs))
   IFMAP(false, xs, ys) -> LAST(xs)
   LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))
   DROPLAST(cons(x, cons(y, xs))) -> DROPLAST(cons(y, xs))

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))
   maplog(xs) -> mapIter(xs, nil)
   mapIter(xs, ys) -> ifmap(isempty(xs), xs, ys)
   ifmap(true, xs, ys) -> ys
   ifmap(false, xs, ys) -> mapIter(droplast(xs), cons(log(last(xs)), ys))
   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   a -> b
   a -> c

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 8 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   DROPLAST(cons(x, cons(y, xs))) -> DROPLAST(cons(y, xs))

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))
   maplog(xs) -> mapIter(xs, nil)
   mapIter(xs, ys) -> ifmap(isempty(xs), xs, ys)
   ifmap(true, xs, ys) -> ys
   ifmap(false, xs, ys) -> mapIter(droplast(xs), cons(log(last(xs)), ys))
   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   a -> b
   a -> c

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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:

   DROPLAST(cons(x, cons(y, xs))) -> DROPLAST(cons(y, xs))

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

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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].

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a


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

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

   DROPLAST(cons(x, cons(y, xs))) -> DROPLAST(cons(y, xs))

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 size-change graphs:
*DROPLAST(cons(x, cons(y, xs))) -> DROPLAST(cons(y, xs))
The graph contains the following edges 1 > 1


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

(13)
YES

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

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

   LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))
   maplog(xs) -> mapIter(xs, nil)
   mapIter(xs, ys) -> ifmap(isempty(xs), xs, ys)
   ifmap(true, xs, ys) -> ys
   ifmap(false, xs, ys) -> mapIter(droplast(xs), cons(log(last(xs)), ys))
   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   a -> b
   a -> c

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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

(15) 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.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))

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

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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

(17) 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].

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a


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

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))

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

(19) 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:
*LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))
The graph contains the following edges 1 > 1


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

(20)
YES

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

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

   DOUBLE(s(x)) -> DOUBLE(x)

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))
   maplog(xs) -> mapIter(xs, nil)
   mapIter(xs, ys) -> ifmap(isempty(xs), xs, ys)
   ifmap(true, xs, ys) -> ys
   ifmap(false, xs, ys) -> mapIter(droplast(xs), cons(log(last(xs)), ys))
   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   a -> b
   a -> c

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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

(22) 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.
----------------------------------------

(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DOUBLE(s(x)) -> DOUBLE(x)

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

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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

(24) 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].

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a


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

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DOUBLE(s(x)) -> DOUBLE(x)

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

(26) 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:
*DOUBLE(s(x)) -> DOUBLE(x)
The graph contains the following edges 1 > 1


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

(27)
YES

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

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

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

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))
   maplog(xs) -> mapIter(xs, nil)
   mapIter(xs, ys) -> ifmap(isempty(xs), xs, ys)
   ifmap(true, xs, ys) -> ys
   ifmap(false, xs, ys) -> mapIter(droplast(xs), cons(log(last(xs)), ys))
   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   a -> b
   a -> c

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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

(29) 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.
----------------------------------------

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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

(31) 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].

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a


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

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

(33) 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:
*LE(s(x), s(y)) -> LE(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(34)
YES

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

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)
   IF(false, x, y, z) -> LOOP(x, double(y), s(z))

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))
   maplog(xs) -> mapIter(xs, nil)
   mapIter(xs, ys) -> ifmap(isempty(xs), xs, ys)
   ifmap(true, xs, ys) -> ys
   ifmap(false, xs, ys) -> mapIter(droplast(xs), cons(log(last(xs)), ys))
   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   a -> b
   a -> c

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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

(36) 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.
----------------------------------------

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)
   IF(false, x, y, z) -> LOOP(x, double(y), s(z))

The TRS R consists of the following rules:

   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   le(s(x), 0) -> false

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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

(38) 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].

   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a


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

(39)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)
   IF(false, x, y, z) -> LOOP(x, double(y), s(z))

The TRS R consists of the following rules:

   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   le(s(x), 0) -> false

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))

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

(40) NonInfProof (EQUIVALENT)
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:

Note that final constraints are written in bold face.



For Pair LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z) the following chains were created:
*We consider the chain LOOP(x3, s(x4), x5) -> IF(le(x3, s(x4)), x3, s(x4), x5), IF(false, x6, x7, x8) -> LOOP(x6, double(x7), s(x8)) which results in the following constraint:

(1)    (IF(le(x3, s(x4)), x3, s(x4), x5)=IF(false, x6, x7, x8)  ==>  LOOP(x3, s(x4), x5)_>=_IF(le(x3, s(x4)), x3, s(x4), x5))



We simplified constraint (1) using rules (I), (II), (IV), (VII) which results in the following new constraint:

(2)    (s(x4)=x18 & le(x3, x18)=false  ==>  LOOP(x3, s(x4), x5)_>=_IF(le(x3, s(x4)), x3, s(x4), x5))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x3, x18)=false which results in the following new constraints:

(3)    (le(x21, x20)=false & s(x4)=s(x20) & (\/x22,x23:le(x21, x20)=false & s(x22)=x20  ==>  LOOP(x21, s(x22), x23)_>=_IF(le(x21, s(x22)), x21, s(x22), x23))  ==>  LOOP(s(x21), s(x4), x5)_>=_IF(le(s(x21), s(x4)), s(x21), s(x4), x5))

(4)    (false=false & s(x4)=0  ==>  LOOP(s(x24), s(x4), x5)_>=_IF(le(s(x24), s(x4)), s(x24), s(x4), x5))



We simplified constraint (3) using rules (I), (II), (III), (IV) which results in the following new constraint:

(5)    (le(x21, x20)=false  ==>  LOOP(s(x21), s(x20), x5)_>=_IF(le(s(x21), s(x20)), s(x21), s(x20), x5))



We solved constraint (4) using rules (I), (II).We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on le(x21, x20)=false which results in the following new constraints:

(6)    (le(x27, x26)=false & (\/x28:le(x27, x26)=false  ==>  LOOP(s(x27), s(x26), x28)_>=_IF(le(s(x27), s(x26)), s(x27), s(x26), x28))  ==>  LOOP(s(s(x27)), s(s(x26)), x5)_>=_IF(le(s(s(x27)), s(s(x26))), s(s(x27)), s(s(x26)), x5))

(7)    (false=false  ==>  LOOP(s(s(x29)), s(0), x5)_>=_IF(le(s(s(x29)), s(0)), s(s(x29)), s(0), x5))



We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (\/x28:le(x27, x26)=false  ==>  LOOP(s(x27), s(x26), x28)_>=_IF(le(s(x27), s(x26)), s(x27), s(x26), x28)) with sigma = [x28 / x5] which results in the following new constraint:

(8)    (LOOP(s(x27), s(x26), x5)_>=_IF(le(s(x27), s(x26)), s(x27), s(x26), x5)  ==>  LOOP(s(s(x27)), s(s(x26)), x5)_>=_IF(le(s(s(x27)), s(s(x26))), s(s(x27)), s(s(x26)), x5))



We simplified constraint (7) using rules (I), (II) which results in the following new constraint:

(9)    (LOOP(s(s(x29)), s(0), x5)_>=_IF(le(s(s(x29)), s(0)), s(s(x29)), s(0), x5))








For Pair IF(false, x, y, z) -> LOOP(x, double(y), s(z)) the following chains were created:
*We consider the chain IF(false, x9, x10, x11) -> LOOP(x9, double(x10), s(x11)), LOOP(x12, s(x13), x14) -> IF(le(x12, s(x13)), x12, s(x13), x14) which results in the following constraint:

(1)    (LOOP(x9, double(x10), s(x11))=LOOP(x12, s(x13), x14)  ==>  IF(false, x9, x10, x11)_>=_LOOP(x9, double(x10), s(x11)))



We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:

(2)    (double(x10)=s(x13)  ==>  IF(false, x9, x10, x11)_>=_LOOP(x9, double(x10), s(x11)))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on double(x10)=s(x13) which results in the following new constraint:

(3)    (s(s(double(x30)))=s(x13) & (\/x31,x32,x33:double(x30)=s(x31)  ==>  IF(false, x32, x30, x33)_>=_LOOP(x32, double(x30), s(x33)))  ==>  IF(false, x9, s(x30), x11)_>=_LOOP(x9, double(s(x30)), s(x11)))



We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:

(4)    (IF(false, x9, s(x30), x11)_>=_LOOP(x9, double(s(x30)), s(x11)))








To summarize, we get the following constraints P__>=_ for the following pairs.

*LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)

*(LOOP(s(x27), s(x26), x5)_>=_IF(le(s(x27), s(x26)), s(x27), s(x26), x5)  ==>  LOOP(s(s(x27)), s(s(x26)), x5)_>=_IF(le(s(s(x27)), s(s(x26))), s(s(x27)), s(s(x26)), x5))


*(LOOP(s(s(x29)), s(0), x5)_>=_IF(le(s(s(x29)), s(0)), s(s(x29)), s(0), x5))




*IF(false, x, y, z) -> LOOP(x, double(y), s(z))

*(IF(false, x9, s(x30), x11)_>=_LOOP(x9, double(s(x30)), s(x11)))








The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by  "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. 

Using the following integer polynomial ordering the  resulting constraints can be solved 

Polynomial interpretation [NONINF]:

   POL(0) = 0
   POL(IF(x_1, x_2, x_3, x_4)) = -1 - x_1 + x_2 - x_3
   POL(LOOP(x_1, x_2, x_3)) = -1 + x_1 - x_2
   POL(c) = -2
   POL(double(x_1)) = 2*x_1
   POL(false) = 0
   POL(le(x_1, x_2)) = 0
   POL(s(x_1)) = 1 + x_1
   POL(true) = 2


The following pairs  are in P_>:
   IF(false, x, y, z) -> LOOP(x, double(y), s(z))
The following pairs are in P_bound:
   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)
The following rules are usable:
   true -> le(0, y)
   le(x, y) -> le(s(x), s(y))
   0 -> double(0)
   s(s(double(x))) -> double(s(x))
   false -> le(s(x), 0)

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

(41)
Complex Obligation (AND)

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

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LOOP(x, s(y), z) -> IF(le(x, s(y)), x, s(y), z)

The TRS R consists of the following rules:

   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   le(s(x), 0) -> false

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))

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

(43) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(44)
TRUE

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

(45)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, x, y, z) -> LOOP(x, double(y), s(z))

The TRS R consists of the following rules:

   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   le(s(x), 0) -> false

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))

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

(46) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(47)
TRUE

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

(48)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFMAP(false, xs, ys) -> MAPITER(droplast(xs), cons(log(last(xs)), ys))
   MAPITER(xs, ys) -> IFMAP(isempty(xs), xs, ys)

The TRS R consists of the following rules:

   le(s(x), 0) -> false
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(true, x, y, z) -> z
   if(false, x, y, z) -> loop(x, double(y), s(z))
   maplog(xs) -> mapIter(xs, nil)
   mapIter(xs, ys) -> ifmap(isempty(xs), xs, ys)
   ifmap(true, xs, ys) -> ys
   ifmap(false, xs, ys) -> mapIter(droplast(xs), cons(log(last(xs)), ys))
   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   a -> b
   a -> c

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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

(49) 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.
----------------------------------------

(50)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFMAP(false, xs, ys) -> MAPITER(droplast(xs), cons(log(last(xs)), ys))
   MAPITER(xs, ys) -> IFMAP(isempty(xs), xs, ys)

The TRS R consists of the following rules:

   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   if(false, x, y, z) -> loop(x, double(y), s(z))
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y, z) -> z
   le(s(x), 0) -> false

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))
   a

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

(51) 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].

   maplog(x0)
   mapIter(x0, x1)
   ifmap(true, x0, x1)
   ifmap(false, x0, x1)
   a


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

(52)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFMAP(false, xs, ys) -> MAPITER(droplast(xs), cons(log(last(xs)), ys))
   MAPITER(xs, ys) -> IFMAP(isempty(xs), xs, ys)

The TRS R consists of the following rules:

   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   if(false, x, y, z) -> loop(x, double(y), s(z))
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y, z) -> z
   le(s(x), 0) -> false

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))

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

(53) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule MAPITER(xs, ys) -> IFMAP(isempty(xs), xs, ys) at position [0] we obtained the following new rules [LPAR04]:

   (MAPITER(nil, y1) -> IFMAP(true, nil, y1),MAPITER(nil, y1) -> IFMAP(true, nil, y1))
   (MAPITER(cons(x0, x1), y1) -> IFMAP(false, cons(x0, x1), y1),MAPITER(cons(x0, x1), y1) -> IFMAP(false, cons(x0, x1), y1))


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

(54)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFMAP(false, xs, ys) -> MAPITER(droplast(xs), cons(log(last(xs)), ys))
   MAPITER(nil, y1) -> IFMAP(true, nil, y1)
   MAPITER(cons(x0, x1), y1) -> IFMAP(false, cons(x0, x1), y1)

The TRS R consists of the following rules:

   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   if(false, x, y, z) -> loop(x, double(y), s(z))
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y, z) -> z
   le(s(x), 0) -> false

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))

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

(55) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(56)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MAPITER(cons(x0, x1), y1) -> IFMAP(false, cons(x0, x1), y1)
   IFMAP(false, xs, ys) -> MAPITER(droplast(xs), cons(log(last(xs)), ys))

The TRS R consists of the following rules:

   isempty(nil) -> true
   isempty(cons(x, xs)) -> false
   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   if(false, x, y, z) -> loop(x, double(y), s(z))
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y, z) -> z
   le(s(x), 0) -> false

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))

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

(57) 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.
----------------------------------------

(58)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MAPITER(cons(x0, x1), y1) -> IFMAP(false, cons(x0, x1), y1)
   IFMAP(false, xs, ys) -> MAPITER(droplast(xs), cons(log(last(xs)), ys))

The TRS R consists of the following rules:

   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(false, x, y, z) -> loop(x, double(y), s(z))
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y, z) -> z
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   le(s(x), 0) -> false

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   isempty(nil)
   isempty(cons(x0, x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))

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

(59) 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].

   isempty(nil)
   isempty(cons(x0, x1))


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

(60)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MAPITER(cons(x0, x1), y1) -> IFMAP(false, cons(x0, x1), y1)
   IFMAP(false, xs, ys) -> MAPITER(droplast(xs), cons(log(last(xs)), ys))

The TRS R consists of the following rules:

   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(false, x, y, z) -> loop(x, double(y), s(z))
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y, z) -> z
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   le(s(x), 0) -> false

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))

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

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


The following pairs can be oriented strictly and are deleted.

   MAPITER(cons(x0, x1), y1) -> IFMAP(false, cons(x0, x1), y1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

   <<<
 POL(MAPITER(x_1, x_2)) =  	[[0]] 	 +  	[[0, 1]] 	* 	x_1 	 +  	[[0, 0]] 	* 	x_2
>>>

   <<<
 POL(cons(x_1, x_2)) =  	[[0], [1]] 	 +  	[[0, 0], [0, 0]] 	* 	x_1 	 +  	[[0, 1], [0, 1]] 	* 	x_2
>>>

   <<<
 POL(IFMAP(x_1, x_2, x_3)) =  	[[0]] 	 +  	[[1, 1]] 	* 	x_1 	 +  	[[1, 0]] 	* 	x_2 	 +  	[[0, 0]] 	* 	x_3
>>>

   <<<
 POL(false) =  	[[0], [0]]
>>>

   <<<
 POL(droplast(x_1)) =  	[[0], [0]] 	 +  	[[1, 0], [1, 0]] 	* 	x_1
>>>

   <<<
 POL(log(x_1)) =  	[[0], [0]] 	 +  	[[0, 0], [0, 0]] 	* 	x_1
>>>

   <<<
 POL(last(x_1)) =  	[[0], [0]] 	 +  	[[0, 0], [0, 0]] 	* 	x_1
>>>

   <<<
 POL(nil) =  	[[1], [1]]
>>>

   <<<
 POL(error) =  	[[0], [0]]
>>>

   <<<
 POL(0) =  	[[0], [0]]
>>>

   <<<
 POL(logError) =  	[[0], [0]]
>>>

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

   <<<
 POL(loop(x_1, x_2, x_3)) =  	[[0], [0]] 	 +  	[[1, 1], [1, 1]] 	* 	x_1 	 +  	[[1, 0], [0, 0]] 	* 	x_2 	 +  	[[1, 1], [1, 1]] 	* 	x_3
>>>

   <<<
 POL(if(x_1, x_2, x_3, x_4)) =  	[[0], [1]] 	 +  	[[0, 0], [0, 0]] 	* 	x_1 	 +  	[[1, 1], [1, 1]] 	* 	x_2 	 +  	[[0, 0], [0, 0]] 	* 	x_3 	 +  	[[1, 1], [0, 1]] 	* 	x_4
>>>

   <<<
 POL(double(x_1)) =  	[[0], [0]] 	 +  	[[0, 0], [1, 0]] 	* 	x_1
>>>

   <<<
 POL(le(x_1, x_2)) =  	[[0], [0]] 	 +  	[[0, 0], [0, 0]] 	* 	x_1 	 +  	[[0, 0], [0, 0]] 	* 	x_2
>>>

   <<<
 POL(true) =  	[[0], [0]]
>>>


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

   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))


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

(62)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFMAP(false, xs, ys) -> MAPITER(droplast(xs), cons(log(last(xs)), ys))

The TRS R consists of the following rules:

   droplast(nil) -> nil
   droplast(cons(x, nil)) -> nil
   droplast(cons(x, cons(y, xs))) -> cons(x, droplast(cons(y, xs)))
   last(nil) -> error
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   log(0) -> logError
   log(s(x)) -> loop(s(x), s(0), 0)
   loop(x, s(y), z) -> if(le(x, s(y)), x, s(y), z)
   if(false, x, y, z) -> loop(x, double(y), s(z))
   le(0, y) -> true
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y, z) -> z
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   le(s(x), 0) -> false

The set Q consists of the following terms:

   le(s(x0), 0)
   le(0, x0)
   le(s(x0), s(x1))
   double(0)
   double(s(x0))
   log(0)
   log(s(x0))
   loop(x0, s(x1), x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   droplast(nil)
   droplast(cons(x0, nil))
   droplast(cons(x0, cons(x1, x2)))

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

(63) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(64)
TRUE