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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 22 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) QDPOrderProof [EQUIVALENT, 22 ms]
        (20) QDP
        (21) DependencyGraphProof [EQUIVALENT, 0 ms]
        (22) TRUE
    (23) QDP
        (24) UsableRulesProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QReductionProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QDPOrderProof [EQUIVALENT, 24 ms]
        (29) QDP
        (30) DependencyGraphProof [EQUIVALENT, 0 ms]
        (31) TRUE
    (32) QDP
        (33) UsableRulesProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) QReductionProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) TransformationProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) TransformationProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) DependencyGraphProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) TransformationProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) UsableRulesProof [EQUIVALENT, 0 ms]
        (46) QDP
        (47) TransformationProof [EQUIVALENT, 0 ms]
        (48) QDP
        (49) TransformationProof [EQUIVALENT, 0 ms]
        (50) QDP
        (51) UsableRulesProof [EQUIVALENT, 0 ms]
        (52) QDP
        (53) TransformationProof [EQUIVALENT, 0 ms]
        (54) QDP
        (55) TransformationProof [EQUIVALENT, 0 ms]
        (56) QDP
        (57) UsableRulesProof [EQUIVALENT, 0 ms]
        (58) QDP
        (59) QReductionProof [EQUIVALENT, 0 ms]
        (60) QDP
        (61) TransformationProof [EQUIVALENT, 0 ms]
        (62) QDP
        (63) TransformationProof [EQUIVALENT, 0 ms]
        (64) QDP
        (65) DependencyGraphProof [EQUIVALENT, 0 ms]
        (66) QDP
        (67) UsableRulesProof [EQUIVALENT, 0 ms]
        (68) QDP
        (69) QReductionProof [EQUIVALENT, 0 ms]
        (70) QDP
        (71) TransformationProof [EQUIVALENT, 0 ms]
        (72) QDP
        (73) DependencyGraphProof [EQUIVALENT, 0 ms]
        (74) QDP
        (75) UsableRulesProof [EQUIVALENT, 0 ms]
        (76) QDP
        (77) QReductionProof [EQUIVALENT, 0 ms]
        (78) QDP
        (79) TransformationProof [EQUIVALENT, 0 ms]
        (80) QDP
        (81) UsableRulesProof [EQUIVALENT, 0 ms]
        (82) QDP
        (83) QReductionProof [EQUIVALENT, 0 ms]
        (84) QDP
        (85) TransformationProof [EQUIVALENT, 0 ms]
        (86) QDP
        (87) QDPOrderProof [EQUIVALENT, 10 ms]
        (88) QDP
        (89) DependencyGraphProof [EQUIVALENT, 0 ms]
        (90) TRUE


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

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

   car(cons(x, l)) -> x
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   cadr(cons(x, cons(y, l))) -> y
   isZero(0) -> true
   isZero(s(x)) -> false
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   times(x, y) -> iftimes(isZero(x), x, y)
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   prod(l) -> if(shorter(l, 0), shorter(l, s(0)), l)
   if(true, b, l) -> s(0)
   if(false, b, l) -> if2(b, l)
   if2(true, l) -> car(l)
   if2(false, l) -> prod(cons(times(car(l), cadr(l)), cddr(l)))

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:

   car(cons(x, l)) -> x
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   cadr(cons(x, cons(y, l))) -> y
   isZero(0) -> true
   isZero(s(x)) -> false
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   times(x, y) -> iftimes(isZero(x), x, y)
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   prod(l) -> if(shorter(l, 0), shorter(l, s(0)), l)
   if(true, b, l) -> s(0)
   if(false, b, l) -> if2(b, l)
   if2(true, l) -> car(l)
   if2(false, l) -> prod(cons(times(car(l), cadr(l)), cddr(l)))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)


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

(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:

   PLUS(x, y) -> IFPLUS(isZero(x), x, y)
   PLUS(x, y) -> ISZERO(x)
   IFPLUS(false, x, y) -> PLUS(p(x), y)
   IFPLUS(false, x, y) -> P(x)
   TIMES(x, y) -> IFTIMES(isZero(x), x, y)
   TIMES(x, y) -> ISZERO(x)
   IFTIMES(false, x, y) -> PLUS(y, times(p(x), y))
   IFTIMES(false, x, y) -> TIMES(p(x), y)
   IFTIMES(false, x, y) -> P(x)
   SHORTER(cons(x, l), s(y)) -> SHORTER(l, y)
   PROD(l) -> IF(shorter(l, 0), shorter(l, s(0)), l)
   PROD(l) -> SHORTER(l, 0)
   PROD(l) -> SHORTER(l, s(0))
   IF(false, b, l) -> IF2(b, l)
   IF2(true, l) -> CAR(l)
   IF2(false, l) -> PROD(cons(times(car(l), cadr(l)), cddr(l)))
   IF2(false, l) -> TIMES(car(l), cadr(l))
   IF2(false, l) -> CAR(l)
   IF2(false, l) -> CADR(l)
   IF2(false, l) -> CDDR(l)

The TRS R consists of the following rules:

   car(cons(x, l)) -> x
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   cadr(cons(x, cons(y, l))) -> y
   isZero(0) -> true
   isZero(s(x)) -> false
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   times(x, y) -> iftimes(isZero(x), x, y)
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   prod(l) -> if(shorter(l, 0), shorter(l, s(0)), l)
   if(true, b, l) -> s(0)
   if(false, b, l) -> if2(b, l)
   if2(true, l) -> car(l)
   if2(false, l) -> prod(cons(times(car(l), cadr(l)), cddr(l)))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)

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 12 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   SHORTER(cons(x, l), s(y)) -> SHORTER(l, y)

The TRS R consists of the following rules:

   car(cons(x, l)) -> x
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   cadr(cons(x, cons(y, l))) -> y
   isZero(0) -> true
   isZero(s(x)) -> false
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   times(x, y) -> iftimes(isZero(x), x, y)
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   prod(l) -> if(shorter(l, 0), shorter(l, s(0)), l)
   if(true, b, l) -> s(0)
   if(false, b, l) -> if2(b, l)
   if2(true, l) -> car(l)
   if2(false, l) -> prod(cons(times(car(l), cadr(l)), cddr(l)))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)

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:

   SHORTER(cons(x, l), s(y)) -> SHORTER(l, y)

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

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)

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].

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)


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

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

   SHORTER(cons(x, l), s(y)) -> SHORTER(l, 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 size-change graphs:
*SHORTER(cons(x, l), s(y)) -> SHORTER(l, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(13)
YES

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

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

   IFPLUS(false, x, y) -> PLUS(p(x), y)
   PLUS(x, y) -> IFPLUS(isZero(x), x, y)

The TRS R consists of the following rules:

   car(cons(x, l)) -> x
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   cadr(cons(x, cons(y, l))) -> y
   isZero(0) -> true
   isZero(s(x)) -> false
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   times(x, y) -> iftimes(isZero(x), x, y)
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   prod(l) -> if(shorter(l, 0), shorter(l, s(0)), l)
   if(true, b, l) -> s(0)
   if(false, b, l) -> if2(b, l)
   if2(true, l) -> car(l)
   if2(false, l) -> prod(cons(times(car(l), cadr(l)), cddr(l)))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)

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:

   IFPLUS(false, x, y) -> PLUS(p(x), y)
   PLUS(x, y) -> IFPLUS(isZero(x), x, y)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)

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].

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)


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

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

   IFPLUS(false, x, y) -> PLUS(p(x), y)
   PLUS(x, y) -> IFPLUS(isZero(x), x, y)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   p(s(x0))
   p(0)

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

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


The following pairs can be oriented strictly and are deleted.

   IFPLUS(false, x, y) -> PLUS(p(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(0) = 0
   POL(IFPLUS(x_1, x_2, x_3)) = [1/4]x_1 + [1/4]x_2
   POL(PLUS(x_1, x_2)) = x_1
   POL(false) = [1/4]
   POL(isZero(x_1)) = [1/2]x_1
   POL(p(x_1)) = [1/4]x_1
   POL(s(x_1)) = [2] + [4]x_1
   POL(true) = 0
The value of delta used in the strict ordering is 1/16.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   p(s(x)) -> x
   p(0) -> 0
   isZero(0) -> true
   isZero(s(x)) -> false


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

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PLUS(x, y) -> IFPLUS(isZero(x), x, y)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   p(s(x0))
   p(0)

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

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

(22)
TRUE

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

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

   IFTIMES(false, x, y) -> TIMES(p(x), y)
   TIMES(x, y) -> IFTIMES(isZero(x), x, y)

The TRS R consists of the following rules:

   car(cons(x, l)) -> x
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   cadr(cons(x, cons(y, l))) -> y
   isZero(0) -> true
   isZero(s(x)) -> false
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   times(x, y) -> iftimes(isZero(x), x, y)
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   prod(l) -> if(shorter(l, 0), shorter(l, s(0)), l)
   if(true, b, l) -> s(0)
   if(false, b, l) -> if2(b, l)
   if2(true, l) -> car(l)
   if2(false, l) -> prod(cons(times(car(l), cadr(l)), cddr(l)))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)

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

(24) 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.
----------------------------------------

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

   IFTIMES(false, x, y) -> TIMES(p(x), y)
   TIMES(x, y) -> IFTIMES(isZero(x), x, y)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)

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

(26) 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].

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)


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

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IFTIMES(false, x, y) -> TIMES(p(x), y)
   TIMES(x, y) -> IFTIMES(isZero(x), x, y)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   p(s(x0))
   p(0)

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

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


The following pairs can be oriented strictly and are deleted.

   IFTIMES(false, x, y) -> TIMES(p(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(0) = 0
   POL(IFTIMES(x_1, x_2, x_3)) = [1/4]x_1 + [1/4]x_2
   POL(TIMES(x_1, x_2)) = x_1
   POL(false) = [1/4]
   POL(isZero(x_1)) = [1/2]x_1
   POL(p(x_1)) = [1/4]x_1
   POL(s(x_1)) = [2] + [4]x_1
   POL(true) = 0
The value of delta used in the strict ordering is 1/16.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   p(s(x)) -> x
   p(0) -> 0
   isZero(0) -> true
   isZero(s(x)) -> false


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

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

   TIMES(x, y) -> IFTIMES(isZero(x), x, y)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   p(s(x0))
   p(0)

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

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

(31)
TRUE

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

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

   IF2(false, l) -> PROD(cons(times(car(l), cadr(l)), cddr(l)))
   PROD(l) -> IF(shorter(l, 0), shorter(l, s(0)), l)
   IF(false, b, l) -> IF2(b, l)

The TRS R consists of the following rules:

   car(cons(x, l)) -> x
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   cadr(cons(x, cons(y, l))) -> y
   isZero(0) -> true
   isZero(s(x)) -> false
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   times(x, y) -> iftimes(isZero(x), x, y)
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   prod(l) -> if(shorter(l, 0), shorter(l, s(0)), l)
   if(true, b, l) -> s(0)
   if(false, b, l) -> if2(b, l)
   if2(true, l) -> car(l)
   if2(false, l) -> prod(cons(times(car(l), cadr(l)), cddr(l)))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)

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

(33) 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.
----------------------------------------

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, l) -> PROD(cons(times(car(l), cadr(l)), cddr(l)))
   PROD(l) -> IF(shorter(l, 0), shorter(l, s(0)), l)
   IF(false, b, l) -> IF2(b, l)

The TRS R consists of the following rules:

   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   car(cons(x, l)) -> x
   cadr(cons(x, cons(y, l))) -> y
   times(x, y) -> iftimes(isZero(x), x, y)
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   isZero(0) -> true
   isZero(s(x)) -> false
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))
   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)

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

(35) 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].

   prod(x0)
   if(true, x0, x1)
   if(false, x0, x1)
   if2(true, x0)
   if2(false, x0)


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

(36)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, l) -> PROD(cons(times(car(l), cadr(l)), cddr(l)))
   PROD(l) -> IF(shorter(l, 0), shorter(l, s(0)), l)
   IF(false, b, l) -> IF2(b, l)

The TRS R consists of the following rules:

   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   car(cons(x, l)) -> x
   cadr(cons(x, cons(y, l))) -> y
   times(x, y) -> iftimes(isZero(x), x, y)
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   isZero(0) -> true
   isZero(s(x)) -> false
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(37) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule IF2(false, l) -> PROD(cons(times(car(l), cadr(l)), cddr(l))) at position [0,0] we obtained the following new rules [LPAR04]:

   (IF2(false, l) -> PROD(cons(iftimes(isZero(car(l)), car(l), cadr(l)), cddr(l))),IF2(false, l) -> PROD(cons(iftimes(isZero(car(l)), car(l), cadr(l)), cddr(l))))


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

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PROD(l) -> IF(shorter(l, 0), shorter(l, s(0)), l)
   IF(false, b, l) -> IF2(b, l)
   IF2(false, l) -> PROD(cons(iftimes(isZero(car(l)), car(l), cadr(l)), cddr(l)))

The TRS R consists of the following rules:

   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   car(cons(x, l)) -> x
   cadr(cons(x, cons(y, l))) -> y
   times(x, y) -> iftimes(isZero(x), x, y)
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   isZero(0) -> true
   isZero(s(x)) -> false
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(39) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule PROD(l) -> IF(shorter(l, 0), shorter(l, s(0)), l) at position [0] we obtained the following new rules [LPAR04]:

   (PROD(nil) -> IF(true, shorter(nil, s(0)), nil),PROD(nil) -> IF(true, shorter(nil, s(0)), nil))
   (PROD(cons(x0, x1)) -> IF(false, shorter(cons(x0, x1), s(0)), cons(x0, x1)),PROD(cons(x0, x1)) -> IF(false, shorter(cons(x0, x1), s(0)), cons(x0, x1)))


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

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, b, l) -> IF2(b, l)
   IF2(false, l) -> PROD(cons(iftimes(isZero(car(l)), car(l), cadr(l)), cddr(l)))
   PROD(nil) -> IF(true, shorter(nil, s(0)), nil)
   PROD(cons(x0, x1)) -> IF(false, shorter(cons(x0, x1), s(0)), cons(x0, x1))

The TRS R consists of the following rules:

   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   car(cons(x, l)) -> x
   cadr(cons(x, cons(y, l))) -> y
   times(x, y) -> iftimes(isZero(x), x, y)
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   isZero(0) -> true
   isZero(s(x)) -> false
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

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

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

   IF2(false, l) -> PROD(cons(iftimes(isZero(car(l)), car(l), cadr(l)), cddr(l)))
   PROD(cons(x0, x1)) -> IF(false, shorter(cons(x0, x1), s(0)), cons(x0, x1))
   IF(false, b, l) -> IF2(b, l)

The TRS R consists of the following rules:

   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   car(cons(x, l)) -> x
   cadr(cons(x, cons(y, l))) -> y
   times(x, y) -> iftimes(isZero(x), x, y)
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   isZero(0) -> true
   isZero(s(x)) -> false
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(43) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule PROD(cons(x0, x1)) -> IF(false, shorter(cons(x0, x1), s(0)), cons(x0, x1)) at position [1] we obtained the following new rules [LPAR04]:

   (PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1)),PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1)))


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

(44)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, l) -> PROD(cons(iftimes(isZero(car(l)), car(l), cadr(l)), cddr(l)))
   IF(false, b, l) -> IF2(b, l)
   PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1))

The TRS R consists of the following rules:

   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   shorter(cons(x, l), s(y)) -> shorter(l, y)
   car(cons(x, l)) -> x
   cadr(cons(x, cons(y, l))) -> y
   times(x, y) -> iftimes(isZero(x), x, y)
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   isZero(0) -> true
   isZero(s(x)) -> false
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(45) 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.
----------------------------------------

(46)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, l) -> PROD(cons(iftimes(isZero(car(l)), car(l), cadr(l)), cddr(l)))
   IF(false, b, l) -> IF2(b, l)
   PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1))

The TRS R consists of the following rules:

   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   car(cons(x, l)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(47) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF(false, b, l) -> IF2(b, l) we obtained the following new rules [LPAR04]:

   (IF(false, y_0, cons(z0, z1)) -> IF2(y_0, cons(z0, z1)),IF(false, y_0, cons(z0, z1)) -> IF2(y_0, cons(z0, z1)))


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

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

   IF2(false, l) -> PROD(cons(iftimes(isZero(car(l)), car(l), cadr(l)), cddr(l)))
   PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1))
   IF(false, y_0, cons(z0, z1)) -> IF2(y_0, cons(z0, z1))

The TRS R consists of the following rules:

   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   car(cons(x, l)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(49) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF2(false, l) -> PROD(cons(iftimes(isZero(car(l)), car(l), cadr(l)), cddr(l))) we obtained the following new rules [LPAR04]:

   (IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(car(cons(z1, z2))), car(cons(z1, z2)), cadr(cons(z1, z2))), cddr(cons(z1, z2)))),IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(car(cons(z1, z2))), car(cons(z1, z2)), cadr(cons(z1, z2))), cddr(cons(z1, z2)))))


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

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

   PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1))
   IF(false, y_0, cons(z0, z1)) -> IF2(y_0, cons(z0, z1))
   IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(car(cons(z1, z2))), car(cons(z1, z2)), cadr(cons(z1, z2))), cddr(cons(z1, z2))))

The TRS R consists of the following rules:

   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false
   car(cons(x, l)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(nil) -> nil
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(51) 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.
----------------------------------------

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

   PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1))
   IF(false, y_0, cons(z0, z1)) -> IF2(y_0, cons(z0, z1))
   IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(car(cons(z1, z2))), car(cons(z1, z2)), cadr(cons(z1, z2))), cddr(cons(z1, z2))))

The TRS R consists of the following rules:

   car(cons(x, l)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(53) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(car(cons(z1, z2))), car(cons(z1, z2)), cadr(cons(z1, z2))), cddr(cons(z1, z2)))) at position [0,0,0,0] we obtained the following new rules [LPAR04]:

   (IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), car(cons(z1, z2)), cadr(cons(z1, z2))), cddr(cons(z1, z2)))),IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), car(cons(z1, z2)), cadr(cons(z1, z2))), cddr(cons(z1, z2)))))


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

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

   PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1))
   IF(false, y_0, cons(z0, z1)) -> IF2(y_0, cons(z0, z1))
   IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), car(cons(z1, z2)), cadr(cons(z1, z2))), cddr(cons(z1, z2))))

The TRS R consists of the following rules:

   car(cons(x, l)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(55) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), car(cons(z1, z2)), cadr(cons(z1, z2))), cddr(cons(z1, z2)))) at position [0,0,1] we obtained the following new rules [LPAR04]:

   (IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), z1, cadr(cons(z1, z2))), cddr(cons(z1, z2)))),IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), z1, cadr(cons(z1, z2))), cddr(cons(z1, z2)))))


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

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

   PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1))
   IF(false, y_0, cons(z0, z1)) -> IF2(y_0, cons(z0, z1))
   IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), z1, cadr(cons(z1, z2))), cddr(cons(z1, z2))))

The TRS R consists of the following rules:

   car(cons(x, l)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(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:

   PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1))
   IF(false, y_0, cons(z0, z1)) -> IF2(y_0, cons(z0, z1))
   IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), z1, cadr(cons(z1, z2))), cddr(cons(z1, z2))))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false

The set Q consists of the following terms:

   car(cons(x0, x1))
   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(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].

   car(cons(x0, x1))


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

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

   PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1))
   IF(false, y_0, cons(z0, z1)) -> IF2(y_0, cons(z0, z1))
   IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), z1, cadr(cons(z1, z2))), cddr(cons(z1, z2))))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false

The set Q consists of the following terms:

   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(61) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule IF(false, y_0, cons(z0, z1)) -> IF2(y_0, cons(z0, z1)) we obtained the following new rules [LPAR04]:

   (IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2)),IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2)))


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

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

   PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1))
   IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), z1, cadr(cons(z1, z2))), cddr(cons(z1, z2))))
   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false

The set Q consists of the following terms:

   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(63) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule PROD(cons(x0, x1)) -> IF(false, shorter(x1, 0), cons(x0, x1)) at position [1] we obtained the following new rules [LPAR04]:

   (PROD(cons(y0, nil)) -> IF(false, true, cons(y0, nil)),PROD(cons(y0, nil)) -> IF(false, true, cons(y0, nil)))
   (PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1))),PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1))))


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

(64)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), z1, cadr(cons(z1, z2))), cddr(cons(z1, z2))))
   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))
   PROD(cons(y0, nil)) -> IF(false, true, cons(y0, nil))
   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false

The set Q consists of the following terms:

   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

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

(66)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))
   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))
   IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), z1, cadr(cons(z1, z2))), cddr(cons(z1, z2))))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))
   shorter(nil, y) -> true
   shorter(cons(x, l), 0) -> false

The set Q consists of the following terms:

   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(67) 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.
----------------------------------------

(68)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))
   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))
   IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), z1, cadr(cons(z1, z2))), cddr(cons(z1, z2))))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)
   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))

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

(69) 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].

   shorter(nil, x0)
   shorter(cons(x0, x1), 0)
   shorter(cons(x0, x1), s(x2))


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

(70)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))
   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))
   IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), z1, cadr(cons(z1, z2))), cddr(cons(z1, z2))))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)

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

(71) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule IF2(false, cons(z1, z2)) -> PROD(cons(iftimes(isZero(z1), z1, cadr(cons(z1, z2))), cddr(cons(z1, z2)))) at position [0,1] we obtained the following new rules [LPAR04]:

   (IF2(false, cons(x0, nil)) -> PROD(cons(iftimes(isZero(x0), x0, cadr(cons(x0, nil))), nil)),IF2(false, cons(x0, nil)) -> PROD(cons(iftimes(isZero(x0), x0, cadr(cons(x0, nil))), nil)))
   (IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, cadr(cons(x0, cons(x1, x2)))), x2)),IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, cadr(cons(x0, cons(x1, x2)))), x2)))


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

(72)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))
   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))
   IF2(false, cons(x0, nil)) -> PROD(cons(iftimes(isZero(x0), x0, cadr(cons(x0, nil))), nil))
   IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, cadr(cons(x0, cons(x1, x2)))), x2))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)

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

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

(74)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))
   IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, cadr(cons(x0, cons(x1, x2)))), x2))
   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   cddr(cons(x, nil)) -> nil
   cddr(cons(x, cons(y, l))) -> l
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)

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

(75) 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.
----------------------------------------

(76)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))
   IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, cadr(cons(x0, cons(x1, x2)))), x2))
   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))
   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)

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

(77) 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].

   cddr(nil)
   cddr(cons(x0, nil))
   cddr(cons(x0, cons(x1, x2)))


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

(78)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))
   IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, cadr(cons(x0, cons(x1, x2)))), x2))
   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)

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

(79) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, cadr(cons(x0, cons(x1, x2)))), x2)) at position [0,0,2] we obtained the following new rules [LPAR04]:

   (IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, x1), x2)),IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, x1), x2)))


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

(80)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))
   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))
   IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, x1), x2))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   cadr(cons(x, cons(y, l))) -> y
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)

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

(81) 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.
----------------------------------------

(82)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))
   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))
   IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, x1), x2))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   cadr(cons(x0, cons(x1, x2)))
   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)

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

(83) 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].

   cadr(cons(x0, cons(x1, x2)))


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

(84)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2))
   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))
   IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, x1), x2))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)

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

(85) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF(false, false, cons(x1, x2)) -> IF2(false, cons(x1, x2)) we obtained the following new rules [LPAR04]:

   (IF(false, false, cons(z0, cons(z1, z2))) -> IF2(false, cons(z0, cons(z1, z2))),IF(false, false, cons(z0, cons(z1, z2))) -> IF2(false, cons(z0, cons(z1, z2))))


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

(86)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))
   IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, x1), x2))
   IF(false, false, cons(z0, cons(z1, z2))) -> IF2(false, cons(z0, cons(z1, z2)))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)

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

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


The following pairs can be oriented strictly and are deleted.

   IF2(false, cons(x0, cons(x1, x2))) -> PROD(cons(iftimes(isZero(x0), x0, x1), x2))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
PROD(x1)  =  x1

cons(x1, x2)  =  cons(x2)

IF(x1, x2, x3)  =  x3

IF2(x1, x2)  =  x2

false  =  false


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   cons_1=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none


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

(88)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PROD(cons(y0, cons(x0, x1))) -> IF(false, false, cons(y0, cons(x0, x1)))
   IF(false, false, cons(z0, cons(z1, z2))) -> IF2(false, cons(z0, cons(z1, z2)))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   iftimes(true, x, y) -> 0
   iftimes(false, x, y) -> plus(y, times(p(x), y))
   p(s(x)) -> x
   p(0) -> 0
   times(x, y) -> iftimes(isZero(x), x, y)
   plus(x, y) -> ifplus(isZero(x), x, y)
   ifplus(true, x, y) -> y
   ifplus(false, x, y) -> s(plus(p(x), y))

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   plus(x0, x1)
   ifplus(true, x0, x1)
   ifplus(false, x0, x1)
   times(x0, x1)
   iftimes(true, x0, x1)
   iftimes(false, x0, x1)
   p(s(x0))
   p(0)

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

(89) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
----------------------------------------

(90)
TRUE