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


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 18 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) QDPOrderProof [EQUIVALENT, 44 ms]
        (13) QDP
        (14) DependencyGraphProof [EQUIVALENT, 0 ms]
        (15) TRUE
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QReductionProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (22) YES
    (23) QDP
        (24) UsableRulesProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QReductionProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) TransformationProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) TransformationProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) TransformationProof [EQUIVALENT, 0 ms]
        (33) QDP
        (34) QDPQMonotonicMRRProof [EQUIVALENT, 17 ms]
        (35) QDP
        (36) TransformationProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) TransformationProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) DependencyGraphProof [EQUIVALENT, 0 ms]
        (41) QDP
        (42) TransformationProof [EQUIVALENT, 0 ms]
        (43) QDP
        (44) TransformationProof [EQUIVALENT, 0 ms]
        (45) QDP
        (46) TransformationProof [EQUIVALENT, 0 ms]
        (47) QDP
        (48) DependencyGraphProof [EQUIVALENT, 0 ms]
        (49) QDP
        (50) TransformationProof [EQUIVALENT, 0 ms]
        (51) QDP
        (52) TransformationProof [EQUIVALENT, 0 ms]
        (53) QDP
        (54) TransformationProof [EQUIVALENT, 0 ms]
        (55) QDP
        (56) TransformationProof [EQUIVALENT, 0 ms]
        (57) QDP
        (58) UsableRulesProof [EQUIVALENT, 0 ms]
        (59) QDP
        (60) QReductionProof [EQUIVALENT, 0 ms]
        (61) QDP
        (62) TransformationProof [EQUIVALENT, 0 ms]
        (63) QDP
        (64) DependencyGraphProof [EQUIVALENT, 0 ms]
        (65) QDP
        (66) UsableRulesProof [EQUIVALENT, 0 ms]
        (67) QDP
        (68) TransformationProof [EQUIVALENT, 0 ms]
        (69) QDP
        (70) TransformationProof [EQUIVALENT, 0 ms]
        (71) QDP
        (72) DependencyGraphProof [EQUIVALENT, 0 ms]
        (73) AND
            (74) QDP
                (75) TransformationProof [EQUIVALENT, 0 ms]
                (76) QDP
                (77) QDPQMonotonicMRRProof [EQUIVALENT, 8 ms]
                (78) QDP
                (79) QDPOrderProof [EQUIVALENT, 243 ms]
                (80) QDP
                (81) DependencyGraphProof [EQUIVALENT, 0 ms]
                (82) TRUE
            (83) QDP
                (84) MRRProof [EQUIVALENT, 2 ms]
                (85) QDP
                (86) QDPOrderProof [EQUIVALENT, 5 ms]
                (87) QDP
                (88) DependencyGraphProof [EQUIVALENT, 0 ms]
                (89) TRUE


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

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

   empty(nil) -> true
   empty(cons(x, y)) -> false
   tail(nil) -> nil
   tail(cons(x, y)) -> y
   head(cons(x, y)) -> x
   zero(0) -> true
   zero(s(x)) -> false
   p(0) -> 0
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   intlist(x) -> if_intlist(empty(x), x)
   if_intlist(true, x) -> nil
   if_intlist(false, x) -> cons(s(head(x)), intlist(tail(x)))
   int(x, y) -> if_int(zero(x), zero(y), x, y)
   if_int(true, b, x, y) -> if1(b, x, y)
   if_int(false, b, x, y) -> if2(b, x, y)
   if1(true, x, y) -> cons(0, nil)
   if1(false, x, y) -> cons(0, int(s(0), y))
   if2(true, x, y) -> nil
   if2(false, x, y) -> intlist(int(p(x), p(y)))

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:

   empty(nil) -> true
   empty(cons(x, y)) -> false
   tail(nil) -> nil
   tail(cons(x, y)) -> y
   head(cons(x, y)) -> x
   zero(0) -> true
   zero(s(x)) -> false
   p(0) -> 0
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   intlist(x) -> if_intlist(empty(x), x)
   if_intlist(true, x) -> nil
   if_intlist(false, x) -> cons(s(head(x)), intlist(tail(x)))
   int(x, y) -> if_int(zero(x), zero(y), x, y)
   if_int(true, b, x, y) -> if1(b, x, y)
   if_int(false, b, x, y) -> if2(b, x, y)
   if1(true, x, y) -> cons(0, nil)
   if1(false, x, y) -> cons(0, int(s(0), y))
   if2(true, x, y) -> nil
   if2(false, x, y) -> intlist(int(p(x), p(y)))

The set Q consists of the following terms:

   empty(nil)
   empty(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))
   intlist(x0)
   if_intlist(true, x0)
   if_intlist(false, x0)
   int(x0, x1)
   if_int(true, x0, x1, x2)
   if_int(false, x0, x1, x2)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)


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

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   P(s(s(x))) -> P(s(x))
   INTLIST(x) -> IF_INTLIST(empty(x), x)
   INTLIST(x) -> EMPTY(x)
   IF_INTLIST(false, x) -> HEAD(x)
   IF_INTLIST(false, x) -> INTLIST(tail(x))
   IF_INTLIST(false, x) -> TAIL(x)
   INT(x, y) -> IF_INT(zero(x), zero(y), x, y)
   INT(x, y) -> ZERO(x)
   INT(x, y) -> ZERO(y)
   IF_INT(true, b, x, y) -> IF1(b, x, y)
   IF_INT(false, b, x, y) -> IF2(b, x, y)
   IF1(false, x, y) -> INT(s(0), y)
   IF2(false, x, y) -> INTLIST(int(p(x), p(y)))
   IF2(false, x, y) -> INT(p(x), p(y))
   IF2(false, x, y) -> P(x)
   IF2(false, x, y) -> P(y)

The TRS R consists of the following rules:

   empty(nil) -> true
   empty(cons(x, y)) -> false
   tail(nil) -> nil
   tail(cons(x, y)) -> y
   head(cons(x, y)) -> x
   zero(0) -> true
   zero(s(x)) -> false
   p(0) -> 0
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   intlist(x) -> if_intlist(empty(x), x)
   if_intlist(true, x) -> nil
   if_intlist(false, x) -> cons(s(head(x)), intlist(tail(x)))
   int(x, y) -> if_int(zero(x), zero(y), x, y)
   if_int(true, b, x, y) -> if1(b, x, y)
   if_int(false, b, x, y) -> if2(b, x, y)
   if1(true, x, y) -> cons(0, nil)
   if1(false, x, y) -> cons(0, int(s(0), y))
   if2(true, x, y) -> nil
   if2(false, x, y) -> intlist(int(p(x), p(y)))

The set Q consists of the following terms:

   empty(nil)
   empty(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))
   intlist(x0)
   if_intlist(true, x0)
   if_intlist(false, x0)
   int(x0, x1)
   if_int(true, x0, x1, x2)
   if_int(false, x0, x1, x2)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 8 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   IF_INTLIST(false, x) -> INTLIST(tail(x))
   INTLIST(x) -> IF_INTLIST(empty(x), x)

The TRS R consists of the following rules:

   empty(nil) -> true
   empty(cons(x, y)) -> false
   tail(nil) -> nil
   tail(cons(x, y)) -> y
   head(cons(x, y)) -> x
   zero(0) -> true
   zero(s(x)) -> false
   p(0) -> 0
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   intlist(x) -> if_intlist(empty(x), x)
   if_intlist(true, x) -> nil
   if_intlist(false, x) -> cons(s(head(x)), intlist(tail(x)))
   int(x, y) -> if_int(zero(x), zero(y), x, y)
   if_int(true, b, x, y) -> if1(b, x, y)
   if_int(false, b, x, y) -> if2(b, x, y)
   if1(true, x, y) -> cons(0, nil)
   if1(false, x, y) -> cons(0, int(s(0), y))
   if2(true, x, y) -> nil
   if2(false, x, y) -> intlist(int(p(x), p(y)))

The set Q consists of the following terms:

   empty(nil)
   empty(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))
   intlist(x0)
   if_intlist(true, x0)
   if_intlist(false, x0)
   int(x0, x1)
   if_int(true, x0, x1, x2)
   if_int(false, x0, x1, x2)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)

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

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

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

   IF_INTLIST(false, x) -> INTLIST(tail(x))
   INTLIST(x) -> IF_INTLIST(empty(x), x)

The TRS R consists of the following rules:

   empty(nil) -> true
   empty(cons(x, y)) -> false
   tail(nil) -> nil
   tail(cons(x, y)) -> y

The set Q consists of the following terms:

   empty(nil)
   empty(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))
   intlist(x0)
   if_intlist(true, x0)
   if_intlist(false, x0)
   int(x0, x1)
   if_int(true, x0, x1, x2)
   if_int(false, x0, x1, x2)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)

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

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

   head(cons(x0, x1))
   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))
   intlist(x0)
   if_intlist(true, x0)
   if_intlist(false, x0)
   int(x0, x1)
   if_int(true, x0, x1, x2)
   if_int(false, x0, x1, x2)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)


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

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

   IF_INTLIST(false, x) -> INTLIST(tail(x))
   INTLIST(x) -> IF_INTLIST(empty(x), x)

The TRS R consists of the following rules:

   empty(nil) -> true
   empty(cons(x, y)) -> false
   tail(nil) -> nil
   tail(cons(x, y)) -> y

The set Q consists of the following terms:

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

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

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


The following pairs can be oriented strictly and are deleted.

   INTLIST(x) -> IF_INTLIST(empty(x), x)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(IF_INTLIST(x_1, x_2)) = [1/4]x_1 + x_2
   POL(INTLIST(x_1)) = [1/4] + [2]x_1
   POL(cons(x_1, x_2)) = [1] + [4]x_2
   POL(empty(x_1)) = [1/4] + [2]x_1
   POL(false) = [1]
   POL(nil) = 0
   POL(tail(x_1)) = [1/2]x_1
   POL(true) = [1/4]
The value of delta used in the strict ordering is 3/16.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   tail(nil) -> nil
   tail(cons(x, y)) -> y
   empty(nil) -> true
   empty(cons(x, y)) -> false


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

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF_INTLIST(false, x) -> INTLIST(tail(x))

The TRS R consists of the following rules:

   empty(nil) -> true
   empty(cons(x, y)) -> false
   tail(nil) -> nil
   tail(cons(x, y)) -> y

The set Q consists of the following terms:

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

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

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

(15)
TRUE

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

(16)
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:

   empty(nil) -> true
   empty(cons(x, y)) -> false
   tail(nil) -> nil
   tail(cons(x, y)) -> y
   head(cons(x, y)) -> x
   zero(0) -> true
   zero(s(x)) -> false
   p(0) -> 0
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   intlist(x) -> if_intlist(empty(x), x)
   if_intlist(true, x) -> nil
   if_intlist(false, x) -> cons(s(head(x)), intlist(tail(x)))
   int(x, y) -> if_int(zero(x), zero(y), x, y)
   if_int(true, b, x, y) -> if1(b, x, y)
   if_int(false, b, x, y) -> if2(b, x, y)
   if1(true, x, y) -> cons(0, nil)
   if1(false, x, y) -> cons(0, int(s(0), y))
   if2(true, x, y) -> nil
   if2(false, x, y) -> intlist(int(p(x), p(y)))

The set Q consists of the following terms:

   empty(nil)
   empty(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))
   intlist(x0)
   if_intlist(true, x0)
   if_intlist(false, x0)
   int(x0, x1)
   if_int(true, x0, x1, x2)
   if_int(false, x0, x1, x2)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)

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

(17) 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.
----------------------------------------

(18)
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:

   empty(nil)
   empty(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))
   intlist(x0)
   if_intlist(true, x0)
   if_intlist(false, x0)
   int(x0, x1)
   if_int(true, x0, x1, x2)
   if_int(false, x0, x1, x2)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)

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

(19) 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].

   empty(nil)
   empty(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))
   intlist(x0)
   if_intlist(true, x0)
   if_intlist(false, x0)
   int(x0, x1)
   if_int(true, x0, x1, x2)
   if_int(false, x0, x1, x2)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)


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

(20)
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.
----------------------------------------

(21) 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


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

(22)
YES

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

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

   IF2(false, x, y) -> INT(p(x), p(y))
   INT(x, y) -> IF_INT(zero(x), zero(y), x, y)
   IF_INT(true, b, x, y) -> IF1(b, x, y)
   IF1(false, x, y) -> INT(s(0), y)
   IF_INT(false, b, x, y) -> IF2(b, x, y)

The TRS R consists of the following rules:

   empty(nil) -> true
   empty(cons(x, y)) -> false
   tail(nil) -> nil
   tail(cons(x, y)) -> y
   head(cons(x, y)) -> x
   zero(0) -> true
   zero(s(x)) -> false
   p(0) -> 0
   p(s(0)) -> 0
   p(s(s(x))) -> s(p(s(x)))
   intlist(x) -> if_intlist(empty(x), x)
   if_intlist(true, x) -> nil
   if_intlist(false, x) -> cons(s(head(x)), intlist(tail(x)))
   int(x, y) -> if_int(zero(x), zero(y), x, y)
   if_int(true, b, x, y) -> if1(b, x, y)
   if_int(false, b, x, y) -> if2(b, x, y)
   if1(true, x, y) -> cons(0, nil)
   if1(false, x, y) -> cons(0, int(s(0), y))
   if2(true, x, y) -> nil
   if2(false, x, y) -> intlist(int(p(x), p(y)))

The set Q consists of the following terms:

   empty(nil)
   empty(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))
   intlist(x0)
   if_intlist(true, x0)
   if_intlist(false, x0)
   int(x0, x1)
   if_int(true, x0, x1, x2)
   if_int(false, x0, x1, x2)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)

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:

   IF2(false, x, y) -> INT(p(x), p(y))
   INT(x, y) -> IF_INT(zero(x), zero(y), x, y)
   IF_INT(true, b, x, y) -> IF1(b, x, y)
   IF1(false, x, y) -> INT(s(0), y)
   IF_INT(false, b, x, y) -> IF2(b, x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   empty(nil)
   empty(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))
   intlist(x0)
   if_intlist(true, x0)
   if_intlist(false, x0)
   int(x0, x1)
   if_int(true, x0, x1, x2)
   if_int(false, x0, x1, x2)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)

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].

   empty(nil)
   empty(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   head(cons(x0, x1))
   intlist(x0)
   if_intlist(true, x0)
   if_intlist(false, x0)
   int(x0, x1)
   if_int(true, x0, x1, x2)
   if_int(false, x0, x1, x2)
   if1(true, x0, x1)
   if1(false, x0, x1)
   if2(true, x0, x1)
   if2(false, x0, x1)


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

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

   IF2(false, x, y) -> INT(p(x), p(y))
   INT(x, y) -> IF_INT(zero(x), zero(y), x, y)
   IF_INT(true, b, x, y) -> IF1(b, x, y)
   IF1(false, x, y) -> INT(s(0), y)
   IF_INT(false, b, x, y) -> IF2(b, x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(28) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule INT(x, y) -> IF_INT(zero(x), zero(y), x, y) at position [0] we obtained the following new rules [LPAR04]:

   (INT(0, y1) -> IF_INT(true, zero(y1), 0, y1),INT(0, y1) -> IF_INT(true, zero(y1), 0, y1))
   (INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1),INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1))


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

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

   IF2(false, x, y) -> INT(p(x), p(y))
   IF_INT(true, b, x, y) -> IF1(b, x, y)
   IF1(false, x, y) -> INT(s(0), y)
   IF_INT(false, b, x, y) -> IF2(b, x, y)
   INT(0, y1) -> IF_INT(true, zero(y1), 0, y1)
   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(30) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF_INT(true, b, x, y) -> IF1(b, x, y) we obtained the following new rules [LPAR04]:

   (IF_INT(true, y_0, 0, z0) -> IF1(y_0, 0, z0),IF_INT(true, y_0, 0, z0) -> IF1(y_0, 0, z0))


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

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, x, y) -> INT(p(x), p(y))
   IF1(false, x, y) -> INT(s(0), y)
   IF_INT(false, b, x, y) -> IF2(b, x, y)
   INT(0, y1) -> IF_INT(true, zero(y1), 0, y1)
   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(true, y_0, 0, z0) -> IF1(y_0, 0, z0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(32) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF1(false, x, y) -> INT(s(0), y) we obtained the following new rules [LPAR04]:

   (IF1(false, 0, z1) -> INT(s(0), z1),IF1(false, 0, z1) -> INT(s(0), z1))


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

(33)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, x, y) -> INT(p(x), p(y))
   IF_INT(false, b, x, y) -> IF2(b, x, y)
   INT(0, y1) -> IF_INT(true, zero(y1), 0, y1)
   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(true, y_0, 0, z0) -> IF1(y_0, 0, z0)
   IF1(false, 0, z1) -> INT(s(0), z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(34) 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 rules of the TRS R:

   zero(0) -> true

Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 1
   POL(IF1(x_1, x_2, x_3)) = 2 + 2*x_1
   POL(IF2(x_1, x_2, x_3)) = 2*x_1 + 2*x_2
   POL(IF_INT(x_1, x_2, x_3, x_4)) = 2*x_2 + 2*x_3
   POL(INT(x_1, x_2)) = 2 + 2*x_1
   POL(false) = 1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 1
   POL(true) = 0
   POL(zero(x_1)) = 1


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

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

   IF2(false, x, y) -> INT(p(x), p(y))
   IF_INT(false, b, x, y) -> IF2(b, x, y)
   INT(0, y1) -> IF_INT(true, zero(y1), 0, y1)
   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(true, y_0, 0, z0) -> IF1(y_0, 0, z0)
   IF1(false, 0, z1) -> INT(s(0), z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(36) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule IF2(false, x, y) -> INT(p(x), p(y)) at position [0] we obtained the following new rules [LPAR04]:

   (IF2(false, 0, y1) -> INT(0, p(y1)),IF2(false, 0, y1) -> INT(0, p(y1)))
   (IF2(false, s(0), y1) -> INT(0, p(y1)),IF2(false, s(0), y1) -> INT(0, p(y1)))
   (IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1)),IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1)))


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

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

   IF_INT(false, b, x, y) -> IF2(b, x, y)
   INT(0, y1) -> IF_INT(true, zero(y1), 0, y1)
   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(true, y_0, 0, z0) -> IF1(y_0, 0, z0)
   IF1(false, 0, z1) -> INT(s(0), z1)
   IF2(false, 0, y1) -> INT(0, p(y1))
   IF2(false, s(0), y1) -> INT(0, p(y1))
   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(38) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF_INT(false, b, x, y) -> IF2(b, x, y) we obtained the following new rules [LPAR04]:

   (IF_INT(false, y_0, s(z0), z1) -> IF2(y_0, s(z0), z1),IF_INT(false, y_0, s(z0), z1) -> IF2(y_0, s(z0), z1))


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

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

   INT(0, y1) -> IF_INT(true, zero(y1), 0, y1)
   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(true, y_0, 0, z0) -> IF1(y_0, 0, z0)
   IF1(false, 0, z1) -> INT(s(0), z1)
   IF2(false, 0, y1) -> INT(0, p(y1))
   IF2(false, s(0), y1) -> INT(0, p(y1))
   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))
   IF_INT(false, y_0, s(z0), z1) -> IF2(y_0, s(z0), z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

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

(41)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF_INT(true, y_0, 0, z0) -> IF1(y_0, 0, z0)
   IF1(false, 0, z1) -> INT(s(0), z1)
   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(false, y_0, s(z0), z1) -> IF2(y_0, s(z0), z1)
   IF2(false, s(0), y1) -> INT(0, p(y1))
   INT(0, y1) -> IF_INT(true, zero(y1), 0, y1)
   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(42) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule IF_INT(true, y_0, 0, z0) -> IF1(y_0, 0, z0) we obtained the following new rules [LPAR04]:

   (IF_INT(true, false, 0, x1) -> IF1(false, 0, x1),IF_INT(true, false, 0, x1) -> IF1(false, 0, x1))


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

(43)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF1(false, 0, z1) -> INT(s(0), z1)
   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(false, y_0, s(z0), z1) -> IF2(y_0, s(z0), z1)
   IF2(false, s(0), y1) -> INT(0, p(y1))
   INT(0, y1) -> IF_INT(true, zero(y1), 0, y1)
   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))
   IF_INT(true, false, 0, x1) -> IF1(false, 0, x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(44) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule INT(0, y1) -> IF_INT(true, zero(y1), 0, y1) at position [1] we obtained the following new rules [LPAR04]:

   (INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0)),INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0)))


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

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

   IF1(false, 0, z1) -> INT(s(0), z1)
   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(false, y_0, s(z0), z1) -> IF2(y_0, s(z0), z1)
   IF2(false, s(0), y1) -> INT(0, p(y1))
   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))
   IF_INT(true, false, 0, x1) -> IF1(false, 0, x1)
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(46) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule IF2(false, s(0), y1) -> INT(0, p(y1)) at position [1] we obtained the following new rules [LPAR04]:

   (IF2(false, s(0), 0) -> INT(0, 0),IF2(false, s(0), 0) -> INT(0, 0))
   (IF2(false, s(0), s(0)) -> INT(0, 0),IF2(false, s(0), s(0)) -> INT(0, 0))
   (IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0)))),IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0)))))


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

(47)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF1(false, 0, z1) -> INT(s(0), z1)
   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(false, y_0, s(z0), z1) -> IF2(y_0, s(z0), z1)
   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))
   IF_INT(true, false, 0, x1) -> IF1(false, 0, x1)
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF2(false, s(0), 0) -> INT(0, 0)
   IF2(false, s(0), s(0)) -> INT(0, 0)
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(48) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(49)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(false, y_0, s(z0), z1) -> IF2(y_0, s(z0), z1)
   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF_INT(true, false, 0, x1) -> IF1(false, 0, x1)
   IF1(false, 0, z1) -> INT(s(0), z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(50) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF_INT(true, false, 0, x1) -> IF1(false, 0, x1) we obtained the following new rules [LPAR04]:

   (IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0)),IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0)))


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

(51)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(false, y_0, s(z0), z1) -> IF2(y_0, s(z0), z1)
   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF1(false, 0, z1) -> INT(s(0), z1)
   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(52) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF1(false, 0, z1) -> INT(s(0), z1) we obtained the following new rules [LPAR04]:

   (IF1(false, 0, s(z0)) -> INT(s(0), s(z0)),IF1(false, 0, s(z0)) -> INT(s(0), s(z0)))


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

(53)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF_INT(false, y_0, s(z0), z1) -> IF2(y_0, s(z0), z1)
   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   IF1(false, 0, s(z0)) -> INT(s(0), s(z0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

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

   (IF_INT(false, false, s(s(y_0)), x2) -> IF2(false, s(s(y_0)), x2),IF_INT(false, false, s(s(y_0)), x2) -> IF2(false, s(s(y_0)), x2))
   (IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0))),IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0))))


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

(55)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1)
   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   IF1(false, 0, s(z0)) -> INT(s(0), s(z0))
   IF_INT(false, false, s(s(y_0)), x2) -> IF2(false, s(s(y_0)), x2)
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(56) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule INT(s(x0), y1) -> IF_INT(false, zero(y1), s(x0), y1) at position [1] we obtained the following new rules [LPAR04]:

   (INT(s(y0), s(x0)) -> IF_INT(false, false, s(y0), s(x0)),INT(s(y0), s(x0)) -> IF_INT(false, false, s(y0), s(x0)))


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

(57)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   IF1(false, 0, s(z0)) -> INT(s(0), s(z0))
   IF_INT(false, false, s(s(y_0)), x2) -> IF2(false, s(s(y_0)), x2)
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   INT(s(y0), s(x0)) -> IF_INT(false, false, s(y0), s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(58) 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.
----------------------------------------

(59)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   IF1(false, 0, s(z0)) -> INT(s(0), s(z0))
   IF_INT(false, false, s(s(y_0)), x2) -> IF2(false, s(s(y_0)), x2)
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   INT(s(y0), s(x0)) -> IF_INT(false, false, s(y0), s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   zero(0)
   zero(s(x0))
   p(0)
   p(s(0))
   p(s(s(x0)))

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

(60) 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].

   zero(0)
   zero(s(x0))


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

(61)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   IF1(false, 0, s(z0)) -> INT(s(0), s(z0))
   IF_INT(false, false, s(s(y_0)), x2) -> IF2(false, s(s(y_0)), x2)
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   INT(s(y0), s(x0)) -> IF_INT(false, false, s(y0), s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(62) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule IF2(false, s(s(x0)), y1) -> INT(s(p(s(x0))), p(y1)) at position [1] we obtained the following new rules [LPAR04]:

   (IF2(false, s(s(y0)), s(0)) -> INT(s(p(s(y0))), 0),IF2(false, s(s(y0)), s(0)) -> INT(s(p(s(y0))), 0))
   (IF2(false, s(s(y0)), s(s(x0))) -> INT(s(p(s(y0))), s(p(s(x0)))),IF2(false, s(s(y0)), s(s(x0))) -> INT(s(p(s(y0))), s(p(s(x0)))))
   (IF2(false, s(s(y0)), 0) -> INT(s(p(s(y0))), 0),IF2(false, s(s(y0)), 0) -> INT(s(p(s(y0))), 0))


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

(63)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   IF1(false, 0, s(z0)) -> INT(s(0), s(z0))
   IF_INT(false, false, s(s(y_0)), x2) -> IF2(false, s(s(y_0)), x2)
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   INT(s(y0), s(x0)) -> IF_INT(false, false, s(y0), s(x0))
   IF2(false, s(s(y0)), s(0)) -> INT(s(p(s(y0))), 0)
   IF2(false, s(s(y0)), s(s(x0))) -> INT(s(p(s(y0))), s(p(s(x0))))
   IF2(false, s(s(y0)), 0) -> INT(s(p(s(y0))), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(64) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(65)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   IF1(false, 0, s(z0)) -> INT(s(0), s(z0))
   INT(s(y0), s(x0)) -> IF_INT(false, false, s(y0), s(x0))
   IF_INT(false, false, s(s(y_0)), x2) -> IF2(false, s(s(y_0)), x2)
   IF2(false, s(s(y0)), s(s(x0))) -> INT(s(p(s(y0))), s(p(s(x0))))
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(66) 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.
----------------------------------------

(67)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   IF1(false, 0, s(z0)) -> INT(s(0), s(z0))
   INT(s(y0), s(x0)) -> IF_INT(false, false, s(y0), s(x0))
   IF_INT(false, false, s(s(y_0)), x2) -> IF2(false, s(s(y_0)), x2)
   IF2(false, s(s(y0)), s(s(x0))) -> INT(s(p(s(y0))), s(p(s(x0))))
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(68) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF_INT(false, false, s(s(y_0)), x2) -> IF2(false, s(s(y_0)), x2) we obtained the following new rules [LPAR04]:

   (IF_INT(false, false, s(s(x0)), s(z1)) -> IF2(false, s(s(x0)), s(z1)),IF_INT(false, false, s(s(x0)), s(z1)) -> IF2(false, s(s(x0)), s(z1)))


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

(69)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   IF1(false, 0, s(z0)) -> INT(s(0), s(z0))
   INT(s(y0), s(x0)) -> IF_INT(false, false, s(y0), s(x0))
   IF2(false, s(s(y0)), s(s(x0))) -> INT(s(p(s(y0))), s(p(s(x0))))
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   IF_INT(false, false, s(s(x0)), s(z1)) -> IF2(false, s(s(x0)), s(z1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(70) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule INT(s(y0), s(x0)) -> IF_INT(false, false, s(y0), s(x0)) we obtained the following new rules [LPAR04]:

   (INT(s(0), s(s(y_0))) -> IF_INT(false, false, s(0), s(s(y_0))),INT(s(0), s(s(y_0))) -> IF_INT(false, false, s(0), s(s(y_0))))
   (INT(s(s(y_0)), s(x1)) -> IF_INT(false, false, s(s(y_0)), s(x1)),INT(s(s(y_0)), s(x1)) -> IF_INT(false, false, s(s(y_0)), s(x1)))


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

(71)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   IF1(false, 0, s(z0)) -> INT(s(0), s(z0))
   IF2(false, s(s(y0)), s(s(x0))) -> INT(s(p(s(y0))), s(p(s(x0))))
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   IF_INT(false, false, s(s(x0)), s(z1)) -> IF2(false, s(s(x0)), s(z1))
   INT(s(0), s(s(y_0))) -> IF_INT(false, false, s(0), s(s(y_0)))
   INT(s(s(y_0)), s(x1)) -> IF_INT(false, false, s(s(y_0)), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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

(73)
Complex Obligation (AND)

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

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

   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   IF1(false, 0, s(z0)) -> INT(s(0), s(z0))
   INT(s(0), s(s(y_0))) -> IF_INT(false, false, s(0), s(s(y_0)))
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(75) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule IF1(false, 0, s(z0)) -> INT(s(0), s(z0)) we obtained the following new rules [LPAR04]:

   (IF1(false, 0, s(s(y_0))) -> INT(s(0), s(s(y_0))),IF1(false, 0, s(s(y_0))) -> INT(s(0), s(s(y_0))))


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

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

   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   INT(s(0), s(s(y_0))) -> IF_INT(false, false, s(0), s(s(y_0)))
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF1(false, 0, s(s(y_0))) -> INT(s(0), s(s(y_0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(77) 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 rules of the TRS R:

   p(s(0)) -> 0

Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 2
   POL(IF1(x_1, x_2, x_3)) = x_1 + 2*x_3
   POL(IF2(x_1, x_2, x_3)) = 2*x_1 + 2*x_3
   POL(IF_INT(x_1, x_2, x_3, x_4)) = 2*x_2 + 2*x_4
   POL(INT(x_1, x_2)) = 2*x_2
   POL(false) = 0
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 2*x_1
   POL(true) = 2


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

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

   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   INT(s(0), s(s(y_0))) -> IF_INT(false, false, s(0), s(s(y_0)))
   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF1(false, 0, s(s(y_0))) -> INT(s(0), s(s(y_0)))

The TRS R consists of the following rules:

   p(s(s(x))) -> s(p(s(x)))

The set Q consists of the following terms:

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

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

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


The following pairs can be oriented strictly and are deleted.

   IF_INT(false, false, s(0), s(s(y_0))) -> IF2(false, s(0), s(s(y_0)))
   IF2(false, s(0), s(s(x0))) -> INT(0, s(p(s(x0))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( INT_2(x_1, x_2) ) = 2x_2 + 2
POL( s_1(x_1) ) = 2x_1 + 2
POL( p_1(x_1) ) = max{0, x_1 - 2}
POL( IF_INT_4(x_1, ..., x_4) ) = 2x_2 + 2x_4 + 2
POL( true ) = 2
POL( false ) = 0
POL( 0 ) = 2
POL( IF1_3(x_1, ..., x_3) ) = 2x_1 + x_2 + 2x_3
POL( IF2_3(x_1, ..., x_3) ) = 2x_1 + 2x_3 + 1

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

   p(s(s(x))) -> s(p(s(x)))


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

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

   IF_INT(true, false, 0, s(z0)) -> IF1(false, 0, s(z0))
   INT(s(0), s(s(y_0))) -> IF_INT(false, false, s(0), s(s(y_0)))
   INT(0, s(x0)) -> IF_INT(true, false, 0, s(x0))
   IF1(false, 0, s(s(y_0))) -> INT(s(0), s(s(y_0)))

The TRS R consists of the following rules:

   p(s(s(x))) -> s(p(s(x)))

The set Q consists of the following terms:

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

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

(81) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 4 less nodes.
----------------------------------------

(82)
TRUE

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

(83)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INT(s(s(y_0)), s(x1)) -> IF_INT(false, false, s(s(y_0)), s(x1))
   IF_INT(false, false, s(s(x0)), s(z1)) -> IF2(false, s(s(x0)), s(z1))
   IF2(false, s(s(y0)), s(s(x0))) -> INT(s(p(s(y0))), s(p(s(x0))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(84) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.


Strictly oriented rules of the TRS R:

   p(s(0)) -> 0

Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 2
   POL(IF2(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(IF_INT(x_1, x_2, x_3, x_4)) = 2*x_1 + 2*x_2 + x_3 + x_4
   POL(INT(x_1, x_2)) = x_1 + x_2
   POL(false) = 0
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 2*x_1


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

(85)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INT(s(s(y_0)), s(x1)) -> IF_INT(false, false, s(s(y_0)), s(x1))
   IF_INT(false, false, s(s(x0)), s(z1)) -> IF2(false, s(s(x0)), s(z1))
   IF2(false, s(s(y0)), s(s(x0))) -> INT(s(p(s(y0))), s(p(s(x0))))

The TRS R consists of the following rules:

   p(s(s(x))) -> s(p(s(x)))

The set Q consists of the following terms:

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

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

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


The following pairs can be oriented strictly and are deleted.

   IF2(false, s(s(y0)), s(s(x0))) -> INT(s(p(s(y0))), s(p(s(x0))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( INT_2(x_1, x_2) ) = x_1
POL( s_1(x_1) ) = 2x_1 + 2
POL( p_1(x_1) ) = max{0, x_1 - 1}
POL( IF_INT_4(x_1, ..., x_4) ) = 2x_1 + 2x_2 + x_3
POL( false ) = 0
POL( IF2_3(x_1, ..., x_3) ) = 2x_1 + x_2

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

   p(s(s(x))) -> s(p(s(x)))


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

(87)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INT(s(s(y_0)), s(x1)) -> IF_INT(false, false, s(s(y_0)), s(x1))
   IF_INT(false, false, s(s(x0)), s(z1)) -> IF2(false, s(s(x0)), s(z1))

The TRS R consists of the following rules:

   p(s(s(x))) -> s(p(s(x)))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
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(88) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
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(89)
TRUE