/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) 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) TransformationProof [EQUIVALENT, 1 ms]
        (34) QDP
        (35) DependencyGraphProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) UsableRulesProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) QReductionProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) TransformationProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) TransformationProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) DependencyGraphProof [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) UsableRulesProof [EQUIVALENT, 0 ms]
        (56) QDP
        (57) QReductionProof [EQUIVALENT, 0 ms]
        (58) QDP
        (59) TransformationProof [EQUIVALENT, 0 ms]
        (60) QDP
        (61) UsableRulesProof [EQUIVALENT, 0 ms]
        (62) QDP
        (63) QReductionProof [EQUIVALENT, 0 ms]
        (64) QDP
        (65) TransformationProof [EQUIVALENT, 0 ms]
        (66) QDP
        (67) TransformationProof [EQUIVALENT, 0 ms]
        (68) QDP
        (69) TransformationProof [EQUIVALENT, 0 ms]
        (70) QDP
        (71) UsableRulesProof [EQUIVALENT, 0 ms]
        (72) QDP
        (73) TransformationProof [EQUIVALENT, 0 ms]
        (74) QDP
        (75) UsableRulesProof [EQUIVALENT, 0 ms]
        (76) QDP
        (77) TransformationProof [EQUIVALENT, 0 ms]
        (78) QDP
        (79) DependencyGraphProof [EQUIVALENT, 0 ms]
        (80) QDP
        (81) TransformationProof [EQUIVALENT, 0 ms]
        (82) QDP
        (83) TransformationProof [EQUIVALENT, 0 ms]
        (84) QDP
        (85) TransformationProof [EQUIVALENT, 0 ms]
        (86) QDP
        (87) TransformationProof [EQUIVALENT, 0 ms]
        (88) QDP
        (89) QDPQMonotonicMRRProof [EQUIVALENT, 38 ms]
        (90) QDP
        (91) DependencyGraphProof [EQUIVALENT, 0 ms]
        (92) AND
            (93) QDP
                (94) QDPOrderProof [EQUIVALENT, 42 ms]
                (95) QDP
                (96) DependencyGraphProof [EQUIVALENT, 0 ms]
                (97) TRUE
            (98) QDP
                (99) UsableRulesProof [EQUIVALENT, 0 ms]
                (100) QDP
                (101) QReductionProof [EQUIVALENT, 0 ms]
                (102) QDP
                (103) QDPSizeChangeProof [EQUIVALENT, 0 ms]
                (104) YES
    (105) QDP
        (106) UsableRulesProof [EQUIVALENT, 0 ms]
        (107) QDP
        (108) QReductionProof [EQUIVALENT, 0 ms]
        (109) QDP
        (110) NonInfProof [EQUIVALENT, 50 ms]
        (111) QDP
        (112) DependencyGraphProof [EQUIVALENT, 0 ms]
        (113) TRUE


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

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

   times(x, y) -> sum(generate(x, y))
   generate(x, y) -> gen(x, y, 0)
   gen(x, y, z) -> if(ge(z, x), x, y, z)
   if(true, x, y, z) -> nil
   if(false, x, y, z) -> cons(y, gen(x, y, s(z)))
   sum(xs) -> sum2(xs, 0)
   sum2(xs, y) -> ifsum(isNil(xs), isZero(head(xs)), xs, y)
   ifsum(true, b, xs, y) -> y
   ifsum(false, b, xs, y) -> ifsum2(b, xs, y)
   ifsum2(true, xs, y) -> sum2(tail(xs), y)
   ifsum2(false, xs, y) -> sum2(cons(p(head(xs)), tail(xs)), s(y))
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   head(cons(x, xs)) -> x
   head(nil) -> error
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   a -> c
   a -> d

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   generate(x, y) -> gen(x, y, 0)
   gen(x, y, z) -> if(ge(z, x), x, y, z)
   if(true, x, y, z) -> nil
   if(false, x, y, z) -> cons(y, gen(x, y, s(z)))
   ifsum(true, b, xs, y) -> y
   ifsum(false, b, xs, y) -> ifsum2(b, xs, y)
   ifsum2(true, xs, y) -> sum2(tail(xs), y)
   ifsum2(false, xs, y) -> sum2(cons(p(head(xs)), tail(xs)), s(y))
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   head(cons(x, xs)) -> x
   head(nil) -> error
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   sum(xs) -> sum2(xs, 0)
   sum2(xs, y) -> ifsum(isNil(xs), isZero(head(xs)), xs, y)
   times(x, y) -> sum(generate(x, y))

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

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

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

   times(x, y) -> sum(generate(x, y))
   generate(x, y) -> gen(x, y, 0)
   gen(x, y, z) -> if(ge(z, x), x, y, z)
   if(true, x, y, z) -> nil
   if(false, x, y, z) -> cons(y, gen(x, y, s(z)))
   sum(xs) -> sum2(xs, 0)
   sum2(xs, y) -> ifsum(isNil(xs), isZero(head(xs)), xs, y)
   ifsum(true, b, xs, y) -> y
   ifsum(false, b, xs, y) -> ifsum2(b, xs, y)
   ifsum2(true, xs, y) -> sum2(tail(xs), y)
   ifsum2(false, xs, y) -> sum2(cons(p(head(xs)), tail(xs)), s(y))
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   head(cons(x, xs)) -> x
   head(nil) -> error
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   a -> c
   a -> d

The set Q consists of the following terms:

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   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:

   TIMES(x, y) -> SUM(generate(x, y))
   TIMES(x, y) -> GENERATE(x, y)
   GENERATE(x, y) -> GEN(x, y, 0)
   GEN(x, y, z) -> IF(ge(z, x), x, y, z)
   GEN(x, y, z) -> GE(z, x)
   IF(false, x, y, z) -> GEN(x, y, s(z))
   SUM(xs) -> SUM2(xs, 0)
   SUM2(xs, y) -> IFSUM(isNil(xs), isZero(head(xs)), xs, y)
   SUM2(xs, y) -> ISNIL(xs)
   SUM2(xs, y) -> ISZERO(head(xs))
   SUM2(xs, y) -> HEAD(xs)
   IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y)
   IFSUM2(true, xs, y) -> SUM2(tail(xs), y)
   IFSUM2(true, xs, y) -> TAIL(xs)
   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))
   IFSUM2(false, xs, y) -> P(head(xs))
   IFSUM2(false, xs, y) -> HEAD(xs)
   IFSUM2(false, xs, y) -> TAIL(xs)
   ISZERO(s(s(x))) -> ISZERO(s(x))
   P(s(s(x))) -> P(s(x))
   GE(s(x), s(y)) -> GE(x, y)

The TRS R consists of the following rules:

   times(x, y) -> sum(generate(x, y))
   generate(x, y) -> gen(x, y, 0)
   gen(x, y, z) -> if(ge(z, x), x, y, z)
   if(true, x, y, z) -> nil
   if(false, x, y, z) -> cons(y, gen(x, y, s(z)))
   sum(xs) -> sum2(xs, 0)
   sum2(xs, y) -> ifsum(isNil(xs), isZero(head(xs)), xs, y)
   ifsum(true, b, xs, y) -> y
   ifsum(false, b, xs, y) -> ifsum2(b, xs, y)
   ifsum2(true, xs, y) -> sum2(tail(xs), y)
   ifsum2(false, xs, y) -> sum2(cons(p(head(xs)), tail(xs)), s(y))
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   head(cons(x, xs)) -> x
   head(nil) -> error
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   a -> c
   a -> d

The set Q consists of the following terms:

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   a

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 12 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   times(x, y) -> sum(generate(x, y))
   generate(x, y) -> gen(x, y, 0)
   gen(x, y, z) -> if(ge(z, x), x, y, z)
   if(true, x, y, z) -> nil
   if(false, x, y, z) -> cons(y, gen(x, y, s(z)))
   sum(xs) -> sum2(xs, 0)
   sum2(xs, y) -> ifsum(isNil(xs), isZero(head(xs)), xs, y)
   ifsum(true, b, xs, y) -> y
   ifsum(false, b, xs, y) -> ifsum2(b, xs, y)
   ifsum2(true, xs, y) -> sum2(tail(xs), y)
   ifsum2(false, xs, y) -> sum2(cons(p(head(xs)), tail(xs)), s(y))
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   head(cons(x, xs)) -> x
   head(nil) -> error
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   a -> c
   a -> d

The set Q consists of the following terms:

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   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:

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

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

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   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].

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   a


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

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

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

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

(12) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*GE(s(x), s(y)) -> GE(x, 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:

   P(s(s(x))) -> P(s(x))

The TRS R consists of the following rules:

   times(x, y) -> sum(generate(x, y))
   generate(x, y) -> gen(x, y, 0)
   gen(x, y, z) -> if(ge(z, x), x, y, z)
   if(true, x, y, z) -> nil
   if(false, x, y, z) -> cons(y, gen(x, y, s(z)))
   sum(xs) -> sum2(xs, 0)
   sum2(xs, y) -> ifsum(isNil(xs), isZero(head(xs)), xs, y)
   ifsum(true, b, xs, y) -> y
   ifsum(false, b, xs, y) -> ifsum2(b, xs, y)
   ifsum2(true, xs, y) -> sum2(tail(xs), y)
   ifsum2(false, xs, y) -> sum2(cons(p(head(xs)), tail(xs)), s(y))
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   head(cons(x, xs)) -> x
   head(nil) -> error
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   a -> c
   a -> d

The set Q consists of the following terms:

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   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:

   P(s(s(x))) -> P(s(x))

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

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   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].

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   a


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

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

   P(s(s(x))) -> P(s(x))

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


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

(20)
YES

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

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

   ISZERO(s(s(x))) -> ISZERO(s(x))

The TRS R consists of the following rules:

   times(x, y) -> sum(generate(x, y))
   generate(x, y) -> gen(x, y, 0)
   gen(x, y, z) -> if(ge(z, x), x, y, z)
   if(true, x, y, z) -> nil
   if(false, x, y, z) -> cons(y, gen(x, y, s(z)))
   sum(xs) -> sum2(xs, 0)
   sum2(xs, y) -> ifsum(isNil(xs), isZero(head(xs)), xs, y)
   ifsum(true, b, xs, y) -> y
   ifsum(false, b, xs, y) -> ifsum2(b, xs, y)
   ifsum2(true, xs, y) -> sum2(tail(xs), y)
   ifsum2(false, xs, y) -> sum2(cons(p(head(xs)), tail(xs)), s(y))
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   head(cons(x, xs)) -> x
   head(nil) -> error
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   a -> c
   a -> d

The set Q consists of the following terms:

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   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:

   ISZERO(s(s(x))) -> ISZERO(s(x))

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

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   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].

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   a


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

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

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


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

(27)
YES

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

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

   IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y)
   IFSUM2(true, xs, y) -> SUM2(tail(xs), y)
   SUM2(xs, y) -> IFSUM(isNil(xs), isZero(head(xs)), xs, y)
   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))

The TRS R consists of the following rules:

   times(x, y) -> sum(generate(x, y))
   generate(x, y) -> gen(x, y, 0)
   gen(x, y, z) -> if(ge(z, x), x, y, z)
   if(true, x, y, z) -> nil
   if(false, x, y, z) -> cons(y, gen(x, y, s(z)))
   sum(xs) -> sum2(xs, 0)
   sum2(xs, y) -> ifsum(isNil(xs), isZero(head(xs)), xs, y)
   ifsum(true, b, xs, y) -> y
   ifsum(false, b, xs, y) -> ifsum2(b, xs, y)
   ifsum2(true, xs, y) -> sum2(tail(xs), y)
   ifsum2(false, xs, y) -> sum2(cons(p(head(xs)), tail(xs)), s(y))
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   head(cons(x, xs)) -> x
   head(nil) -> error
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   a -> c
   a -> d

The set Q consists of the following terms:

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   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:

   IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y)
   IFSUM2(true, xs, y) -> SUM2(tail(xs), y)
   SUM2(xs, y) -> IFSUM(isNil(xs), isZero(head(xs)), xs, y)
   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   head(nil) -> error
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   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].

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   a


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

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

   IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y)
   IFSUM2(true, xs, y) -> SUM2(tail(xs), y)
   SUM2(xs, y) -> IFSUM(isNil(xs), isZero(head(xs)), xs, y)
   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   head(nil) -> error
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(33) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule SUM2(xs, y) -> IFSUM(isNil(xs), isZero(head(xs)), xs, y) at position [0] we obtained the following new rules [LPAR04]:

   (SUM2(nil, y1) -> IFSUM(true, isZero(head(nil)), nil, y1),SUM2(nil, y1) -> IFSUM(true, isZero(head(nil)), nil, y1))
   (SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(head(cons(x0, x1))), cons(x0, x1), y1),SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(head(cons(x0, x1))), cons(x0, x1), y1))


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

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

   IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y)
   IFSUM2(true, xs, y) -> SUM2(tail(xs), y)
   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))
   SUM2(nil, y1) -> IFSUM(true, isZero(head(nil)), nil, y1)
   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(head(cons(x0, x1))), cons(x0, x1), y1)

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   head(nil) -> error
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

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

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

   IFSUM2(true, xs, y) -> SUM2(tail(xs), y)
   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(head(cons(x0, x1))), cons(x0, x1), y1)
   IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y)
   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   head(nil) -> error
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(37) 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.
----------------------------------------

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

   IFSUM2(true, xs, y) -> SUM2(tail(xs), y)
   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(head(cons(x0, x1))), cons(x0, x1), y1)
   IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y)
   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   head(nil) -> error
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(39) 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].

   isNil(nil)
   isNil(cons(x0, x1))


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

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

   IFSUM2(true, xs, y) -> SUM2(tail(xs), y)
   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(head(cons(x0, x1))), cons(x0, x1), y1)
   IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y)
   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   head(nil) -> error
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(41) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(head(cons(x0, x1))), cons(x0, x1), y1) at position [1,0] we obtained the following new rules [LPAR04]:

   (SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1),SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1))


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

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

   IFSUM2(true, xs, y) -> SUM2(tail(xs), y)
   IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y)
   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))
   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   head(nil) -> error
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(43) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule IFSUM2(true, xs, y) -> SUM2(tail(xs), y) at position [0] we obtained the following new rules [LPAR04]:

   (IFSUM2(true, nil, y1) -> SUM2(nil, y1),IFSUM2(true, nil, y1) -> SUM2(nil, y1))
   (IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1),IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1))


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

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

   IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y)
   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))
   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM2(true, nil, y1) -> SUM2(nil, y1)
   IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1)

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   head(nil) -> error
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

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

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

   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))
   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y)
   IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1)

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   head(nil) -> error
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(47) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IFSUM(false, b, xs, y) -> IFSUM2(b, xs, y) we obtained the following new rules [LPAR04]:

   (IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2),IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2))


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

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

   IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y))
   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1)
   IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2)

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   head(nil) -> error
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(49) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IFSUM2(false, xs, y) -> SUM2(cons(p(head(xs)), tail(xs)), s(y)) we obtained the following new rules [LPAR04]:

   (IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(head(cons(z1, z2))), tail(cons(z1, z2))), s(z3)),IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(head(cons(z1, z2))), tail(cons(z1, z2))), s(z3)))


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

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

   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1)
   IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2)
   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(head(cons(z1, z2))), tail(cons(z1, z2))), s(z3))

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   head(nil) -> error
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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:

   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1)
   IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2)
   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(head(cons(z1, z2))), tail(cons(z1, z2))), s(z3))

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(53) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(head(cons(z1, z2))), tail(cons(z1, z2))), s(z3)) at position [0,0,0] we obtained the following new rules [LPAR04]:

   (IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), tail(cons(z1, z2))), s(z3)),IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), tail(cons(z1, z2))), s(z3)))


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

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

   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1)
   IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2)
   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), tail(cons(z1, z2))), s(z3))

The TRS R consists of the following rules:

   head(cons(x, xs)) -> x
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(55) 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.
----------------------------------------

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

   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1)
   IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2)
   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), tail(cons(z1, z2))), s(z3))

The TRS R consists of the following rules:

   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(57) 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].

   head(cons(x0, x1))
   head(nil)


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

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

   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1)
   IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2)
   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), tail(cons(z1, z2))), s(z3))

The TRS R consists of the following rules:

   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(59) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), tail(cons(z1, z2))), s(z3)) at position [0,1] we obtained the following new rules [LPAR04]:

   (IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), z2), s(z3)),IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), z2), s(z3)))


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

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

   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1)
   IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2)
   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), z2), s(z3))

The TRS R consists of the following rules:

   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   tail(cons(x, xs)) -> xs
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(61) 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.
----------------------------------------

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

   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1)
   IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2)
   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), z2), s(z3))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   tail(nil)
   tail(cons(x0, x1))
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(63) 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].

   tail(nil)
   tail(cons(x0, x1))


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

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

   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1)
   IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2)
   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), z2), s(z3))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(65) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule IFSUM2(true, cons(x0, x1), y1) -> SUM2(x1, y1) we obtained the following new rules [LPAR04]:

   (IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2),IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2))


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

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

   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2)
   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), z2), s(z3))
   IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(67) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule IFSUM(false, y_0, cons(z0, z1), z2) -> IFSUM2(y_0, cons(z0, z1), z2) we obtained the following new rules [LPAR04]:

   (IFSUM(false, false, cons(x1, x2), x3) -> IFSUM2(false, cons(x1, x2), x3),IFSUM(false, false, cons(x1, x2), x3) -> IFSUM2(false, cons(x1, x2), x3))
   (IFSUM(false, true, cons(x1, cons(y_1, y_2)), x3) -> IFSUM2(true, cons(x1, cons(y_1, y_2)), x3),IFSUM(false, true, cons(x1, cons(y_1, y_2)), x3) -> IFSUM2(true, cons(x1, cons(y_1, y_2)), x3))


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

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

   SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1)
   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), z2), s(z3))
   IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2)
   IFSUM(false, false, cons(x1, x2), x3) -> IFSUM2(false, cons(x1, x2), x3)
   IFSUM(false, true, cons(x1, cons(y_1, y_2)), x3) -> IFSUM2(true, cons(x1, cons(y_1, y_2)), x3)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(69) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule SUM2(cons(x0, x1), y1) -> IFSUM(false, isZero(x0), cons(x0, x1), y1) at position [1] we obtained the following new rules [LPAR04]:

   (SUM2(cons(0, y1), y2) -> IFSUM(false, true, cons(0, y1), y2),SUM2(cons(0, y1), y2) -> IFSUM(false, true, cons(0, y1), y2))
   (SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2),SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2))
   (SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2),SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2))


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

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

   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), z2), s(z3))
   IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2)
   IFSUM(false, false, cons(x1, x2), x3) -> IFSUM2(false, cons(x1, x2), x3)
   IFSUM(false, true, cons(x1, cons(y_1, y_2)), x3) -> IFSUM2(true, cons(x1, cons(y_1, y_2)), x3)
   SUM2(cons(0, y1), y2) -> IFSUM(false, true, cons(0, y1), y2)
   SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2)
   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(71) 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.
----------------------------------------

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

   IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), z2), s(z3))
   IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2)
   IFSUM(false, false, cons(x1, x2), x3) -> IFSUM2(false, cons(x1, x2), x3)
   IFSUM(false, true, cons(x1, cons(y_1, y_2)), x3) -> IFSUM2(true, cons(x1, cons(y_1, y_2)), x3)
   SUM2(cons(0, y1), y2) -> IFSUM(false, true, cons(0, y1), y2)
   SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2)
   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(73) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule IFSUM2(false, cons(z1, z2), z3) -> SUM2(cons(p(z1), z2), s(z3)) at position [0,0] we obtained the following new rules [LPAR04]:

   (IFSUM2(false, cons(0, y1), y2) -> SUM2(cons(s(s(0)), y1), s(y2)),IFSUM2(false, cons(0, y1), y2) -> SUM2(cons(s(s(0)), y1), s(y2)))
   (IFSUM2(false, cons(s(0), y1), y2) -> SUM2(cons(0, y1), s(y2)),IFSUM2(false, cons(s(0), y1), y2) -> SUM2(cons(0, y1), s(y2)))
   (IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2)),IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2)))


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

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

   IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2)
   IFSUM(false, false, cons(x1, x2), x3) -> IFSUM2(false, cons(x1, x2), x3)
   IFSUM(false, true, cons(x1, cons(y_1, y_2)), x3) -> IFSUM2(true, cons(x1, cons(y_1, y_2)), x3)
   SUM2(cons(0, y1), y2) -> IFSUM(false, true, cons(0, y1), y2)
   SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2)
   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)
   IFSUM2(false, cons(0, y1), y2) -> SUM2(cons(s(s(0)), y1), s(y2))
   IFSUM2(false, cons(s(0), y1), y2) -> SUM2(cons(0, y1), s(y2))
   IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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:

   IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2)
   IFSUM(false, false, cons(x1, x2), x3) -> IFSUM2(false, cons(x1, x2), x3)
   IFSUM(false, true, cons(x1, cons(y_1, y_2)), x3) -> IFSUM2(true, cons(x1, cons(y_1, y_2)), x3)
   SUM2(cons(0, y1), y2) -> IFSUM(false, true, cons(0, y1), y2)
   SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2)
   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)
   IFSUM2(false, cons(0, y1), y2) -> SUM2(cons(s(s(0)), y1), s(y2))
   IFSUM2(false, cons(s(0), y1), y2) -> SUM2(cons(0, y1), s(y2))
   IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(77) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IFSUM(false, false, cons(x1, x2), x3) -> IFSUM2(false, cons(x1, x2), x3) we obtained the following new rules [LPAR04]:

   (IFSUM(false, false, cons(s(0), z0), z1) -> IFSUM2(false, cons(s(0), z0), z1),IFSUM(false, false, cons(s(0), z0), z1) -> IFSUM2(false, cons(s(0), z0), z1))
   (IFSUM(false, false, cons(s(s(z0)), z1), z2) -> IFSUM2(false, cons(s(s(z0)), z1), z2),IFSUM(false, false, cons(s(s(z0)), z1), z2) -> IFSUM2(false, cons(s(s(z0)), z1), z2))


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

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

   IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2)
   IFSUM(false, true, cons(x1, cons(y_1, y_2)), x3) -> IFSUM2(true, cons(x1, cons(y_1, y_2)), x3)
   SUM2(cons(0, y1), y2) -> IFSUM(false, true, cons(0, y1), y2)
   SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2)
   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)
   IFSUM2(false, cons(0, y1), y2) -> SUM2(cons(s(s(0)), y1), s(y2))
   IFSUM2(false, cons(s(0), y1), y2) -> SUM2(cons(0, y1), s(y2))
   IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2))
   IFSUM(false, false, cons(s(0), z0), z1) -> IFSUM2(false, cons(s(0), z0), z1)
   IFSUM(false, false, cons(s(s(z0)), z1), z2) -> IFSUM2(false, cons(s(s(z0)), z1), z2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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

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

   SUM2(cons(0, y1), y2) -> IFSUM(false, true, cons(0, y1), y2)
   IFSUM(false, true, cons(x1, cons(y_1, y_2)), x3) -> IFSUM2(true, cons(x1, cons(y_1, y_2)), x3)
   IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2)
   SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2)
   IFSUM(false, false, cons(s(0), z0), z1) -> IFSUM2(false, cons(s(0), z0), z1)
   IFSUM2(false, cons(s(0), y1), y2) -> SUM2(cons(0, y1), s(y2))
   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)
   IFSUM(false, false, cons(s(s(z0)), z1), z2) -> IFSUM2(false, cons(s(s(z0)), z1), z2)
   IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(81) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IFSUM(false, true, cons(x1, cons(y_1, y_2)), x3) -> IFSUM2(true, cons(x1, cons(y_1, y_2)), x3) we obtained the following new rules [LPAR04]:

   (IFSUM(false, true, cons(0, cons(x1, x2)), z1) -> IFSUM2(true, cons(0, cons(x1, x2)), z1),IFSUM(false, true, cons(0, cons(x1, x2)), z1) -> IFSUM2(true, cons(0, cons(x1, x2)), z1))


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

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

   SUM2(cons(0, y1), y2) -> IFSUM(false, true, cons(0, y1), y2)
   IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2)
   SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2)
   IFSUM(false, false, cons(s(0), z0), z1) -> IFSUM2(false, cons(s(0), z0), z1)
   IFSUM2(false, cons(s(0), y1), y2) -> SUM2(cons(0, y1), s(y2))
   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)
   IFSUM(false, false, cons(s(s(z0)), z1), z2) -> IFSUM2(false, cons(s(s(z0)), z1), z2)
   IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2))
   IFSUM(false, true, cons(0, cons(x1, x2)), z1) -> IFSUM2(true, cons(0, cons(x1, x2)), z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(83) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IFSUM2(true, cons(x0, cons(y_0, y_1)), x2) -> SUM2(cons(y_0, y_1), x2) we obtained the following new rules [LPAR04]:

   (IFSUM2(true, cons(0, cons(z0, z1)), z2) -> SUM2(cons(z0, z1), z2),IFSUM2(true, cons(0, cons(z0, z1)), z2) -> SUM2(cons(z0, z1), z2))


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

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

   SUM2(cons(0, y1), y2) -> IFSUM(false, true, cons(0, y1), y2)
   SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2)
   IFSUM(false, false, cons(s(0), z0), z1) -> IFSUM2(false, cons(s(0), z0), z1)
   IFSUM2(false, cons(s(0), y1), y2) -> SUM2(cons(0, y1), s(y2))
   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)
   IFSUM(false, false, cons(s(s(z0)), z1), z2) -> IFSUM2(false, cons(s(s(z0)), z1), z2)
   IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2))
   IFSUM(false, true, cons(0, cons(x1, x2)), z1) -> IFSUM2(true, cons(0, cons(x1, x2)), z1)
   IFSUM2(true, cons(0, cons(z0, z1)), z2) -> SUM2(cons(z0, z1), z2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(85) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule SUM2(cons(0, y1), y2) -> IFSUM(false, true, cons(0, y1), y2) we obtained the following new rules [LPAR04]:

   (SUM2(cons(0, cons(y_0, y_1)), x1) -> IFSUM(false, true, cons(0, cons(y_0, y_1)), x1),SUM2(cons(0, cons(y_0, y_1)), x1) -> IFSUM(false, true, cons(0, cons(y_0, y_1)), x1))


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

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

   SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2)
   IFSUM(false, false, cons(s(0), z0), z1) -> IFSUM2(false, cons(s(0), z0), z1)
   IFSUM2(false, cons(s(0), y1), y2) -> SUM2(cons(0, y1), s(y2))
   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)
   IFSUM(false, false, cons(s(s(z0)), z1), z2) -> IFSUM2(false, cons(s(s(z0)), z1), z2)
   IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2))
   IFSUM(false, true, cons(0, cons(x1, x2)), z1) -> IFSUM2(true, cons(0, cons(x1, x2)), z1)
   IFSUM2(true, cons(0, cons(z0, z1)), z2) -> SUM2(cons(z0, z1), z2)
   SUM2(cons(0, cons(y_0, y_1)), x1) -> IFSUM(false, true, cons(0, cons(y_0, y_1)), x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(87) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule IFSUM2(false, cons(s(0), y1), y2) -> SUM2(cons(0, y1), s(y2)) we obtained the following new rules [LPAR04]:

   (IFSUM2(false, cons(s(0), cons(y_0, y_1)), x1) -> SUM2(cons(0, cons(y_0, y_1)), s(x1)),IFSUM2(false, cons(s(0), cons(y_0, y_1)), x1) -> SUM2(cons(0, cons(y_0, y_1)), s(x1)))


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

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

   SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2)
   IFSUM(false, false, cons(s(0), z0), z1) -> IFSUM2(false, cons(s(0), z0), z1)
   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)
   IFSUM(false, false, cons(s(s(z0)), z1), z2) -> IFSUM2(false, cons(s(s(z0)), z1), z2)
   IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2))
   IFSUM(false, true, cons(0, cons(x1, x2)), z1) -> IFSUM2(true, cons(0, cons(x1, x2)), z1)
   IFSUM2(true, cons(0, cons(z0, z1)), z2) -> SUM2(cons(z0, z1), z2)
   SUM2(cons(0, cons(y_0, y_1)), x1) -> IFSUM(false, true, cons(0, cons(y_0, y_1)), x1)
   IFSUM2(false, cons(s(0), cons(y_0, y_1)), x1) -> SUM2(cons(0, cons(y_0, y_1)), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(89) QDPQMonotonicMRRProof (EQUIVALENT)
By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.

Strictly oriented dependency pairs:

   IFSUM2(false, cons(s(0), cons(y_0, y_1)), x1) -> SUM2(cons(0, cons(y_0, y_1)), s(x1))


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(IFSUM(x_1, x_2, x_3, x_4)) = 2*x_1 + 2*x_3
   POL(IFSUM2(x_1, x_2, x_3)) = 2*x_2
   POL(SUM2(x_1, x_2)) = 2*x_1
   POL(cons(x_1, x_2)) = 2*x_1 + x_2
   POL(false) = 0
   POL(isZero(x_1)) = 1
   POL(p(x_1)) = 2*x_1
   POL(s(x_1)) = 2
   POL(true) = 0


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

(90)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM2(cons(s(0), y1), y2) -> IFSUM(false, false, cons(s(0), y1), y2)
   IFSUM(false, false, cons(s(0), z0), z1) -> IFSUM2(false, cons(s(0), z0), z1)
   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)
   IFSUM(false, false, cons(s(s(z0)), z1), z2) -> IFSUM2(false, cons(s(s(z0)), z1), z2)
   IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2))
   IFSUM(false, true, cons(0, cons(x1, x2)), z1) -> IFSUM2(true, cons(0, cons(x1, x2)), z1)
   IFSUM2(true, cons(0, cons(z0, z1)), z2) -> SUM2(cons(z0, z1), z2)
   SUM2(cons(0, cons(y_0, y_1)), x1) -> IFSUM(false, true, cons(0, cons(y_0, y_1)), x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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

(92)
Complex Obligation (AND)

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

(93)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)
   IFSUM(false, false, cons(s(s(z0)), z1), z2) -> IFSUM2(false, cons(s(s(z0)), z1), z2)
   IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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


The following pairs can be oriented strictly and are deleted.

   IFSUM2(false, cons(s(s(x0)), y1), y2) -> SUM2(cons(s(p(s(x0))), y1), s(y2))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( IFSUM_4(x_1, ..., x_4) ) = max{0, 2x_1 + 2x_2 + 2x_3 - 2}
POL( isZero_1(x_1) ) = 1
POL( s_1(x_1) ) = 2x_1 + 1
POL( 0 ) = 0
POL( false ) = 0
POL( SUM2_2(x_1, x_2) ) = 2x_1
POL( cons_2(x_1, x_2) ) = 2x_1 + 1
POL( p_1(x_1) ) = max{0, x_1 - 1}
POL( IFSUM2_3(x_1, ..., x_3) ) = max{0, 2x_1 + 2x_2 - 2}

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

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


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

(95)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM2(cons(s(s(x0)), y1), y2) -> IFSUM(false, isZero(s(x0)), cons(s(s(x0)), y1), y2)
   IFSUM(false, false, cons(s(s(z0)), z1), z2) -> IFSUM2(false, cons(s(s(z0)), z1), z2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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

(97)
TRUE

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

(98)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM2(cons(0, cons(y_0, y_1)), x1) -> IFSUM(false, true, cons(0, cons(y_0, y_1)), x1)
   IFSUM(false, true, cons(0, cons(x1, x2)), z1) -> IFSUM2(true, cons(0, cons(x1, x2)), z1)
   IFSUM2(true, cons(0, cons(z0, z1)), z2) -> SUM2(cons(z0, z1), z2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(99) 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.
----------------------------------------

(100)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM2(cons(0, cons(y_0, y_1)), x1) -> IFSUM(false, true, cons(0, cons(y_0, y_1)), x1)
   IFSUM(false, true, cons(0, cons(x1, x2)), z1) -> IFSUM2(true, cons(0, cons(x1, x2)), z1)
   IFSUM2(true, cons(0, cons(z0, z1)), z2) -> SUM2(cons(z0, z1), z2)

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

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

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

(101) 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].

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


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

(102)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM2(cons(0, cons(y_0, y_1)), x1) -> IFSUM(false, true, cons(0, cons(y_0, y_1)), x1)
   IFSUM(false, true, cons(0, cons(x1, x2)), z1) -> IFSUM2(true, cons(0, cons(x1, x2)), z1)
   IFSUM2(true, cons(0, cons(z0, z1)), z2) -> SUM2(cons(z0, z1), z2)

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

(103) 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:
*IFSUM(false, true, cons(0, cons(x1, x2)), z1) -> IFSUM2(true, cons(0, cons(x1, x2)), z1)
The graph contains the following edges 2 >= 1, 3 >= 2, 4 >= 3


*IFSUM2(true, cons(0, cons(z0, z1)), z2) -> SUM2(cons(z0, z1), z2)
The graph contains the following edges 2 > 1, 3 >= 2


*SUM2(cons(0, cons(y_0, y_1)), x1) -> IFSUM(false, true, cons(0, cons(y_0, y_1)), x1)
The graph contains the following edges 1 >= 3, 2 >= 4


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

(104)
YES

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

(105)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, x, y, z) -> GEN(x, y, s(z))
   GEN(x, y, z) -> IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

   times(x, y) -> sum(generate(x, y))
   generate(x, y) -> gen(x, y, 0)
   gen(x, y, z) -> if(ge(z, x), x, y, z)
   if(true, x, y, z) -> nil
   if(false, x, y, z) -> cons(y, gen(x, y, s(z)))
   sum(xs) -> sum2(xs, 0)
   sum2(xs, y) -> ifsum(isNil(xs), isZero(head(xs)), xs, y)
   ifsum(true, b, xs, y) -> y
   ifsum(false, b, xs, y) -> ifsum2(b, xs, y)
   ifsum2(true, xs, y) -> sum2(tail(xs), y)
   ifsum2(false, xs, y) -> sum2(cons(p(head(xs)), tail(xs)), s(y))
   isNil(nil) -> true
   isNil(cons(x, xs)) -> false
   tail(nil) -> nil
   tail(cons(x, xs)) -> xs
   head(cons(x, xs)) -> x
   head(nil) -> error
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   p(0) -> s(s(0))
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   a -> c
   a -> d

The set Q consists of the following terms:

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   a

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

(106) 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.
----------------------------------------

(107)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, x, y, z) -> GEN(x, y, s(z))
   GEN(x, y, z) -> IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)

The set Q consists of the following terms:

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   a

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

(108) 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].

   times(x0, x1)
   generate(x0, x1)
   gen(x0, x1, x2)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   sum(x0)
   sum2(x0, x1)
   ifsum(true, x0, x1, x2)
   ifsum(false, x0, x1, x2)
   ifsum2(true, x0, x1)
   ifsum2(false, x0, x1)
   isNil(nil)
   isNil(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   head(nil)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   p(0)
   p(s(0))
   p(s(s(x0)))
   a


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

(109)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, x, y, z) -> GEN(x, y, s(z))
   GEN(x, y, z) -> IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

(110) 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 IF(false, x, y, z) -> GEN(x, y, s(z)) the following chains were created:
*We consider the chain GEN(x3, x4, x5) -> IF(ge(x5, x3), x3, x4, x5), IF(false, x6, x7, x8) -> GEN(x6, x7, s(x8)) which results in the following constraint:

(1)    (IF(ge(x5, x3), x3, x4, x5)=IF(false, x6, x7, x8)  ==>  IF(false, x6, x7, x8)_>=_GEN(x6, x7, s(x8)))



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

(2)    (ge(x5, x3)=false  ==>  IF(false, x3, x4, x5)_>=_GEN(x3, x4, s(x5)))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x5, x3)=false which results in the following new constraints:

(3)    (false=false  ==>  IF(false, s(x19), x4, 0)_>=_GEN(s(x19), x4, s(0)))

(4)    (ge(x21, x20)=false & (\/x22:ge(x21, x20)=false  ==>  IF(false, x20, x22, x21)_>=_GEN(x20, x22, s(x21)))  ==>  IF(false, s(x20), x4, s(x21))_>=_GEN(s(x20), x4, s(s(x21))))



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

(5)    (IF(false, s(x19), x4, 0)_>=_GEN(s(x19), x4, s(0)))



We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (\/x22:ge(x21, x20)=false  ==>  IF(false, x20, x22, x21)_>=_GEN(x20, x22, s(x21))) with sigma = [x22 / x4] which results in the following new constraint:

(6)    (IF(false, x20, x4, x21)_>=_GEN(x20, x4, s(x21))  ==>  IF(false, s(x20), x4, s(x21))_>=_GEN(s(x20), x4, s(s(x21))))








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

(1)    (GEN(x9, x10, s(x11))=GEN(x12, x13, x14)  ==>  GEN(x12, x13, x14)_>=_IF(ge(x14, x12), x12, x13, x14))



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

(2)    (GEN(x9, x10, s(x11))_>=_IF(ge(s(x11), x9), x9, x10, s(x11)))








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

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

*(IF(false, s(x19), x4, 0)_>=_GEN(s(x19), x4, s(0)))


*(IF(false, x20, x4, x21)_>=_GEN(x20, x4, s(x21))  ==>  IF(false, s(x20), x4, s(x21))_>=_GEN(s(x20), x4, s(s(x21))))




*GEN(x, y, z) -> IF(ge(z, x), x, y, z)

*(GEN(x9, x10, s(x11))_>=_IF(ge(s(x11), x9), x9, x10, 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(GEN(x_1, x_2, x_3)) = -1 + x_1 + x_2 - x_3
   POL(IF(x_1, x_2, x_3, x_4)) = -1 - x_1 + x_2 + x_3 - x_4
   POL(c) = -2
   POL(false) = 0
   POL(ge(x_1, x_2)) = 0
   POL(s(x_1)) = 1 + x_1
   POL(true) = 0


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

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

(111)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GEN(x, y, z) -> IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

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

(113)
TRUE