/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) AAECC Innermost [EQUIVALENT, 25 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) TransformationProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) TransformationProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) TransformationProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) TransformationProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) DependencyGraphProof [EQUIVALENT, 0 ms]
        (21) TRUE
    (22) QDP
        (23) UsableRulesProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) QReductionProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (28) YES
    (29) QDP
        (30) UsableRulesProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) QReductionProof [EQUIVALENT, 0 ms]
        (33) QDP
        (34) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (35) YES
    (36) QDP
        (37) UsableRulesProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) QReductionProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (42) YES
    (43) QDP
        (44) UsableRulesProof [EQUIVALENT, 0 ms]
        (45) QDP
        (46) QReductionProof [EQUIVALENT, 0 ms]
        (47) QDP
        (48) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (49) YES
    (50) QDP
        (51) UsableRulesProof [EQUIVALENT, 0 ms]
        (52) QDP
        (53) QReductionProof [EQUIVALENT, 0 ms]
        (54) QDP
        (55) QDPQMonotonicMRRProof [EQUIVALENT, 24 ms]
        (56) QDP
        (57) QDPQMonotonicMRRProof [EQUIVALENT, 13 ms]
        (58) QDP
        (59) QDPOrderProof [EQUIVALENT, 0 ms]
        (60) QDP
        (61) DependencyGraphProof [EQUIVALENT, 0 ms]
        (62) TRUE
    (63) QDP
        (64) UsableRulesProof [EQUIVALENT, 0 ms]
        (65) QDP
        (66) QReductionProof [EQUIVALENT, 0 ms]
        (67) QDP
        (68) QDPQMonotonicMRRProof [EQUIVALENT, 33 ms]
        (69) QDP
        (70) NonInfProof [EQUIVALENT, 50 ms]
        (71) AND
            (72) QDP
                (73) DependencyGraphProof [EQUIVALENT, 0 ms]
                (74) TRUE
            (75) QDP
                (76) DependencyGraphProof [EQUIVALENT, 0 ms]
                (77) TRUE


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

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

   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   ifPlus(true, x, y) -> p(y)
   ifPlus(false, x, y) -> plus(p(x), y)
   times(x, y) -> timesIter(0, x, y, 0)
   timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
   ifTimes(true, i, x, y, z) -> z
   ifTimes(false, i, x, y, z) -> timesIter(inc(i), x, y, plus(z, y))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)
   f0(0, y, x) -> f1(x, y, x)
   f1(x, y, z) -> f2(x, y, z)
   f2(x, 1, z) -> f0(x, z, z)
   f0(x, y, z) -> d
   f1(x, y, z) -> c

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   ifTimes(true, i, x, y, z) -> z
   ifTimes(false, i, x, y, z) -> timesIter(inc(i), x, y, plus(z, y))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)
   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   ifPlus(true, x, y) -> p(y)
   ifPlus(false, x, y) -> plus(p(x), y)
   times(x, y) -> timesIter(0, x, y, 0)
   timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)

The TRS R 2 is 
   f0(0, y, x) -> f1(x, y, x)
   f1(x, y, z) -> f2(x, y, z)
   f2(x, 1, z) -> f0(x, z, z)
   f0(x, y, z) -> d
   f1(x, y, z) -> c

The signature Sigma is {f0_3, f1_3, f2_3, d, c}
----------------------------------------

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

   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   ifPlus(true, x, y) -> p(y)
   ifPlus(false, x, y) -> plus(p(x), y)
   times(x, y) -> timesIter(0, x, y, 0)
   timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
   ifTimes(true, i, x, y, z) -> z
   ifTimes(false, i, x, y, z) -> timesIter(inc(i), x, y, plus(z, y))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)
   f0(0, y, x) -> f1(x, y, x)
   f1(x, y, z) -> f2(x, y, z)
   f2(x, 1, z) -> f0(x, z, z)
   f0(x, y, z) -> d
   f1(x, y, z) -> c

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)


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

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

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

   PLUS(x, y) -> IFPLUS(isZero(x), x, inc(y))
   PLUS(x, y) -> ISZERO(x)
   PLUS(x, y) -> INC(y)
   IFPLUS(true, x, y) -> P(y)
   IFPLUS(false, x, y) -> PLUS(p(x), y)
   IFPLUS(false, x, y) -> P(x)
   TIMES(x, y) -> TIMESITER(0, x, y, 0)
   TIMESITER(i, x, y, z) -> IFTIMES(ge(i, x), i, x, y, z)
   TIMESITER(i, x, y, z) -> GE(i, x)
   IFTIMES(false, i, x, y, z) -> TIMESITER(inc(i), x, y, plus(z, y))
   IFTIMES(false, i, x, y, z) -> INC(i)
   IFTIMES(false, i, x, y, z) -> PLUS(z, y)
   ISZERO(s(s(x))) -> ISZERO(s(x))
   INC(s(x)) -> INC(x)
   P(s(s(x))) -> P(s(x))
   GE(s(x), s(y)) -> GE(x, y)
   F0(0, y, x) -> F1(x, y, x)
   F1(x, y, z) -> F2(x, y, z)
   F2(x, 1, z) -> F0(x, z, z)

The TRS R consists of the following rules:

   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   ifPlus(true, x, y) -> p(y)
   ifPlus(false, x, y) -> plus(p(x), y)
   times(x, y) -> timesIter(0, x, y, 0)
   timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
   ifTimes(true, i, x, y, z) -> z
   ifTimes(false, i, x, y, z) -> timesIter(inc(i), x, y, plus(z, y))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)
   f0(0, y, x) -> f1(x, y, x)
   f1(x, y, z) -> f2(x, y, z)
   f2(x, 1, z) -> f0(x, z, z)
   f0(x, y, z) -> d
   f1(x, y, z) -> c

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 8 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   F1(x, y, z) -> F2(x, y, z)
   F2(x, 1, z) -> F0(x, z, z)
   F0(0, y, x) -> F1(x, y, x)

The TRS R consists of the following rules:

   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   ifPlus(true, x, y) -> p(y)
   ifPlus(false, x, y) -> plus(p(x), y)
   times(x, y) -> timesIter(0, x, y, 0)
   timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
   ifTimes(true, i, x, y, z) -> z
   ifTimes(false, i, x, y, z) -> timesIter(inc(i), x, y, plus(z, y))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)
   f0(0, y, x) -> f1(x, y, x)
   f1(x, y, z) -> f2(x, y, z)
   f2(x, 1, z) -> f0(x, z, z)
   f0(x, y, z) -> d
   f1(x, y, z) -> c

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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:

   F1(x, y, z) -> F2(x, y, z)
   F2(x, 1, z) -> F0(x, z, z)
   F0(0, y, x) -> F1(x, y, x)

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

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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].

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)


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

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

   F1(x, y, z) -> F2(x, y, z)
   F2(x, 1, z) -> F0(x, z, z)
   F0(0, y, x) -> F1(x, y, x)

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

(12) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F1(x, y, z) -> F2(x, y, z) we obtained the following new rules [LPAR04]:

   (F1(z1, z0, z1) -> F2(z1, z0, z1),F1(z1, z0, z1) -> F2(z1, z0, z1))


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

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

   F2(x, 1, z) -> F0(x, z, z)
   F0(0, y, x) -> F1(x, y, x)
   F1(z1, z0, z1) -> F2(z1, z0, z1)

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

(14) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F2(x, 1, z) -> F0(x, z, z) we obtained the following new rules [LPAR04]:

   (F2(z0, 1, z0) -> F0(z0, z0, z0),F2(z0, 1, z0) -> F0(z0, z0, z0))


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

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F0(0, y, x) -> F1(x, y, x)
   F1(z1, z0, z1) -> F2(z1, z0, z1)
   F2(z0, 1, z0) -> F0(z0, z0, z0)

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

(16) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F0(0, y, x) -> F1(x, y, x) we obtained the following new rules [LPAR04]:

   (F0(0, 0, 0) -> F1(0, 0, 0),F0(0, 0, 0) -> F1(0, 0, 0))


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

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F1(z1, z0, z1) -> F2(z1, z0, z1)
   F2(z0, 1, z0) -> F0(z0, z0, z0)
   F0(0, 0, 0) -> F1(0, 0, 0)

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

(18) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F1(z1, z0, z1) -> F2(z1, z0, z1) we obtained the following new rules [LPAR04]:

   (F1(0, 0, 0) -> F2(0, 0, 0),F1(0, 0, 0) -> F2(0, 0, 0))


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

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F2(z0, 1, z0) -> F0(z0, z0, z0)
   F0(0, 0, 0) -> F1(0, 0, 0)
   F1(0, 0, 0) -> F2(0, 0, 0)

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

(20) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.
----------------------------------------

(21)
TRUE

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

(22)
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:

   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   ifPlus(true, x, y) -> p(y)
   ifPlus(false, x, y) -> plus(p(x), y)
   times(x, y) -> timesIter(0, x, y, 0)
   timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
   ifTimes(true, i, x, y, z) -> z
   ifTimes(false, i, x, y, z) -> timesIter(inc(i), x, y, plus(z, y))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)
   f0(0, y, x) -> f1(x, y, x)
   f1(x, y, z) -> f2(x, y, z)
   f2(x, 1, z) -> f0(x, z, z)
   f0(x, y, z) -> d
   f1(x, y, z) -> c

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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

(23) 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.
----------------------------------------

(24)
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:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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

(25) 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].

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)


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

(26)
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.
----------------------------------------

(27) 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


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

(28)
YES

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

(29)
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:

   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   ifPlus(true, x, y) -> p(y)
   ifPlus(false, x, y) -> plus(p(x), y)
   times(x, y) -> timesIter(0, x, y, 0)
   timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
   ifTimes(true, i, x, y, z) -> z
   ifTimes(false, i, x, y, z) -> timesIter(inc(i), x, y, plus(z, y))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)
   f0(0, y, x) -> f1(x, y, x)
   f1(x, y, z) -> f2(x, y, z)
   f2(x, 1, z) -> f0(x, z, z)
   f0(x, y, z) -> d
   f1(x, y, z) -> c

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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

(30) 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.
----------------------------------------

(31)
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:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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

(32) 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].

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)


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

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

(34) 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


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

(35)
YES

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

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

   INC(s(x)) -> INC(x)

The TRS R consists of the following rules:

   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   ifPlus(true, x, y) -> p(y)
   ifPlus(false, x, y) -> plus(p(x), y)
   times(x, y) -> timesIter(0, x, y, 0)
   timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
   ifTimes(true, i, x, y, z) -> z
   ifTimes(false, i, x, y, z) -> timesIter(inc(i), x, y, plus(z, y))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)
   f0(0, y, x) -> f1(x, y, x)
   f1(x, y, z) -> f2(x, y, z)
   f2(x, 1, z) -> f0(x, z, z)
   f0(x, y, z) -> d
   f1(x, y, z) -> c

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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:

   INC(s(x)) -> INC(x)

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

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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].

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)


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

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

   INC(s(x)) -> INC(x)

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

(41) 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:
*INC(s(x)) -> INC(x)
The graph contains the following edges 1 > 1


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

(42)
YES

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

(43)
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:

   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   ifPlus(true, x, y) -> p(y)
   ifPlus(false, x, y) -> plus(p(x), y)
   times(x, y) -> timesIter(0, x, y, 0)
   timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
   ifTimes(true, i, x, y, z) -> z
   ifTimes(false, i, x, y, z) -> timesIter(inc(i), x, y, plus(z, y))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)
   f0(0, y, x) -> f1(x, y, x)
   f1(x, y, z) -> f2(x, y, z)
   f2(x, 1, z) -> f0(x, z, z)
   f0(x, y, z) -> d
   f1(x, y, z) -> c

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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

(44) 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.
----------------------------------------

(45)
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:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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

(46) 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].

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)


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

(47)
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.
----------------------------------------

(48) 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


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

(49)
YES

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

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

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

The TRS R consists of the following rules:

   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   ifPlus(true, x, y) -> p(y)
   ifPlus(false, x, y) -> plus(p(x), y)
   times(x, y) -> timesIter(0, x, y, 0)
   timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
   ifTimes(true, i, x, y, z) -> z
   ifTimes(false, i, x, y, z) -> timesIter(inc(i), x, y, plus(z, y))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)
   f0(0, y, x) -> f1(x, y, x)
   f1(x, y, z) -> f2(x, y, z)
   f2(x, 1, z) -> f0(x, z, z)
   f0(x, y, z) -> d
   f1(x, y, z) -> c

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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

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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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

(53) 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].

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)


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

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

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

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)))
   inc(x0)
   p(0)
   p(s(x0))

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

(55) 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:

   isZero(0) -> true

Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 1
   POL(IFPLUS(x_1, x_2, x_3)) = x_1
   POL(PLUS(x_1, x_2)) = 2
   POL(false) = 2
   POL(inc(x_1)) = 2*x_1
   POL(isZero(x_1)) = 2
   POL(p(x_1)) = 2*x_1
   POL(s(x_1)) = 2*x_1
   POL(true) = 1


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

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

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

The TRS R consists of the following rules:

   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)))
   inc(x0)
   p(0)
   p(s(x0))

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

(57) 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(x)) -> x

Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(IFPLUS(x_1, x_2, x_3)) = 2*x_2
   POL(PLUS(x_1, x_2)) = 2*x_1
   POL(false) = 0
   POL(inc(x_1)) = 2 + 2*x_1
   POL(isZero(x_1)) = 0
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 2 + 2*x_1


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

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

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

The TRS R consists of the following rules:

   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(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)))
   inc(x0)
   p(0)
   p(s(x0))

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

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


The following pairs can be oriented strictly and are deleted.

   IFPLUS(false, x, y) -> PLUS(p(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

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

   <<<
 POL(false) =  	[[0], [1]]
>>>

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

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

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

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

   <<<
 POL(0) =  	[[1], [0]]
>>>

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


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

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


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

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

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

The TRS R consists of the following rules:

   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(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)))
   inc(x0)
   p(0)
   p(s(x0))

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

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

(62)
TRUE

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

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

   IFTIMES(false, i, x, y, z) -> TIMESITER(inc(i), x, y, plus(z, y))
   TIMESITER(i, x, y, z) -> IFTIMES(ge(i, x), i, x, y, z)

The TRS R consists of the following rules:

   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   ifPlus(true, x, y) -> p(y)
   ifPlus(false, x, y) -> plus(p(x), y)
   times(x, y) -> timesIter(0, x, y, 0)
   timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
   ifTimes(true, i, x, y, z) -> z
   ifTimes(false, i, x, y, z) -> timesIter(inc(i), x, y, plus(z, y))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   p(0) -> 0
   p(s(x)) -> x
   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)
   f0(0, y, x) -> f1(x, y, x)
   f1(x, y, z) -> f2(x, y, z)
   f2(x, 1, z) -> f0(x, z, z)
   f0(x, y, z) -> d
   f1(x, y, z) -> c

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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

(64) 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.
----------------------------------------

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

   IFTIMES(false, i, x, y, z) -> TIMESITER(inc(i), x, y, plus(z, y))
   TIMESITER(i, x, y, z) -> IFTIMES(ge(i, x), i, 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)
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   ifPlus(false, x, y) -> plus(p(x), y)
   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   p(0) -> 0
   p(s(x)) -> x
   p(s(s(x))) -> s(p(s(x)))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   ifPlus(true, x, y) -> p(y)

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   times(x0, x1)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)

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

(66) 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)
   timesIter(x0, x1, x2, x3)
   ifTimes(true, x0, x1, x2, x3)
   ifTimes(false, x0, x1, x2, x3)
   f1(x0, x1, x2)
   f2(x0, 1, x1)
   f0(x0, x1, x2)


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

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

   IFTIMES(false, i, x, y, z) -> TIMESITER(inc(i), x, y, plus(z, y))
   TIMESITER(i, x, y, z) -> IFTIMES(ge(i, x), i, 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)
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   ifPlus(false, x, y) -> plus(p(x), y)
   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   p(0) -> 0
   p(s(x)) -> x
   p(s(s(x))) -> s(p(s(x)))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   ifPlus(true, x, y) -> p(y)

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

(68) 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:

   ge(x, 0) -> true

Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(IFTIMES(x_1, x_2, x_3, x_4, x_5)) = 2*x_1 + 2*x_3 + 2*x_4
   POL(TIMESITER(x_1, x_2, x_3, x_4)) = 2 + 2*x_2 + 2*x_3
   POL(false) = 1
   POL(ge(x_1, x_2)) = 1
   POL(ifPlus(x_1, x_2, x_3)) = 2*x_3
   POL(inc(x_1)) = x_1
   POL(isZero(x_1)) = 2
   POL(p(x_1)) = 2*x_1
   POL(plus(x_1, x_2)) = 2*x_2
   POL(s(x_1)) = x_1
   POL(true) = 0


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

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

   IFTIMES(false, i, x, y, z) -> TIMESITER(inc(i), x, y, plus(z, y))
   TIMESITER(i, x, y, z) -> IFTIMES(ge(i, x), i, x, y, z)

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   ifPlus(false, x, y) -> plus(p(x), y)
   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   p(0) -> 0
   p(s(x)) -> x
   p(s(s(x))) -> s(p(s(x)))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   ifPlus(true, x, y) -> p(y)

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

(70) 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 IFTIMES(false, i, x, y, z) -> TIMESITER(inc(i), x, y, plus(z, y)) the following chains were created:
*We consider the chain TIMESITER(x4, x5, x6, x7) -> IFTIMES(ge(x4, x5), x4, x5, x6, x7), IFTIMES(false, x8, x9, x10, x11) -> TIMESITER(inc(x8), x9, x10, plus(x11, x10)) which results in the following constraint:

(1)    (IFTIMES(ge(x4, x5), x4, x5, x6, x7)=IFTIMES(false, x8, x9, x10, x11)  ==>  IFTIMES(false, x8, x9, x10, x11)_>=_TIMESITER(inc(x8), x9, x10, plus(x11, x10)))



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

(2)    (ge(x4, x5)=false  ==>  IFTIMES(false, x4, x5, x6, x7)_>=_TIMESITER(inc(x4), x5, x6, plus(x7, x6)))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x4, x5)=false which results in the following new constraints:

(3)    (false=false  ==>  IFTIMES(false, 0, s(x24), x6, x7)_>=_TIMESITER(inc(0), s(x24), x6, plus(x7, x6)))

(4)    (ge(x26, x25)=false & (\/x27,x28:ge(x26, x25)=false  ==>  IFTIMES(false, x26, x25, x27, x28)_>=_TIMESITER(inc(x26), x25, x27, plus(x28, x27)))  ==>  IFTIMES(false, s(x26), s(x25), x6, x7)_>=_TIMESITER(inc(s(x26)), s(x25), x6, plus(x7, x6)))



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

(5)    (IFTIMES(false, 0, s(x24), x6, x7)_>=_TIMESITER(inc(0), s(x24), x6, plus(x7, x6)))



We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (\/x27,x28:ge(x26, x25)=false  ==>  IFTIMES(false, x26, x25, x27, x28)_>=_TIMESITER(inc(x26), x25, x27, plus(x28, x27))) with sigma = [x27 / x6, x28 / x7] which results in the following new constraint:

(6)    (IFTIMES(false, x26, x25, x6, x7)_>=_TIMESITER(inc(x26), x25, x6, plus(x7, x6))  ==>  IFTIMES(false, s(x26), s(x25), x6, x7)_>=_TIMESITER(inc(s(x26)), s(x25), x6, plus(x7, x6)))








For Pair TIMESITER(i, x, y, z) -> IFTIMES(ge(i, x), i, x, y, z) the following chains were created:
*We consider the chain IFTIMES(false, x12, x13, x14, x15) -> TIMESITER(inc(x12), x13, x14, plus(x15, x14)), TIMESITER(x16, x17, x18, x19) -> IFTIMES(ge(x16, x17), x16, x17, x18, x19) which results in the following constraint:

(1)    (TIMESITER(inc(x12), x13, x14, plus(x15, x14))=TIMESITER(x16, x17, x18, x19)  ==>  TIMESITER(x16, x17, x18, x19)_>=_IFTIMES(ge(x16, x17), x16, x17, x18, x19))



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

(2)    (TIMESITER(x16, x13, x14, x19)_>=_IFTIMES(ge(x16, x13), x16, x13, x14, x19))








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

*IFTIMES(false, i, x, y, z) -> TIMESITER(inc(i), x, y, plus(z, y))

*(IFTIMES(false, 0, s(x24), x6, x7)_>=_TIMESITER(inc(0), s(x24), x6, plus(x7, x6)))


*(IFTIMES(false, x26, x25, x6, x7)_>=_TIMESITER(inc(x26), x25, x6, plus(x7, x6))  ==>  IFTIMES(false, s(x26), s(x25), x6, x7)_>=_TIMESITER(inc(s(x26)), s(x25), x6, plus(x7, x6)))




*TIMESITER(i, x, y, z) -> IFTIMES(ge(i, x), i, x, y, z)

*(TIMESITER(x16, x13, x14, x19)_>=_IFTIMES(ge(x16, x13), x16, x13, x14, x19))








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(IFTIMES(x_1, x_2, x_3, x_4, x_5)) = -x_1 - x_2 + x_3 + x_4
   POL(TIMESITER(x_1, x_2, x_3, x_4)) = 1 - x_1 + x_2 + x_3
   POL(c) = -1
   POL(false) = 0
   POL(ge(x_1, x_2)) = 0
   POL(ifPlus(x_1, x_2, x_3)) = 0
   POL(inc(x_1)) = 1 + x_1
   POL(isZero(x_1)) = x_1
   POL(p(x_1)) = 1 + x_1
   POL(plus(x_1, x_2)) = x_1 + x_2
   POL(s(x_1)) = 1 + x_1
   POL(true) = 1


The following pairs  are in P_>:
   TIMESITER(i, x, y, z) -> IFTIMES(ge(i, x), i, x, y, z)
The following pairs are in P_bound:
   IFTIMES(false, i, x, y, z) -> TIMESITER(inc(i), x, y, plus(z, y))
The following rules are usable:
   s(0) -> inc(0)
   s(inc(x)) -> inc(s(x))
   s(x) -> inc(x)
   false -> ge(0, s(y))
   ge(x, y) -> ge(s(x), s(y))

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

(71)
Complex Obligation (AND)

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

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

   IFTIMES(false, i, x, y, z) -> TIMESITER(inc(i), x, y, plus(z, y))

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   ifPlus(false, x, y) -> plus(p(x), y)
   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   p(0) -> 0
   p(s(x)) -> x
   p(s(s(x))) -> s(p(s(x)))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   ifPlus(true, x, y) -> p(y)

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

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

(74)
TRUE

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

(75)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TIMESITER(i, x, y, z) -> IFTIMES(ge(i, x), i, x, y, z)

The TRS R consists of the following rules:

   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   inc(0) -> s(0)
   inc(s(x)) -> s(inc(x))
   inc(x) -> s(x)
   ifPlus(false, x, y) -> plus(p(x), y)
   plus(x, y) -> ifPlus(isZero(x), x, inc(y))
   p(0) -> 0
   p(s(x)) -> x
   p(s(s(x))) -> s(p(s(x)))
   isZero(0) -> true
   isZero(s(0)) -> false
   isZero(s(s(x))) -> isZero(s(x))
   ifPlus(true, x, y) -> p(y)

The set Q consists of the following terms:

   plus(x0, x1)
   ifPlus(true, x0, x1)
   ifPlus(false, x0, x1)
   isZero(0)
   isZero(s(0))
   isZero(s(s(x0)))
   inc(x0)
   p(0)
   p(s(x0))
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

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

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

(77)
TRUE