/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 4 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES
    (28) QDP
        (29) UsableRulesProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) QReductionProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) TransformationProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) DependencyGraphProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) UsableRulesProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) QReductionProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) TransformationProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) UsableRulesProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) TransformationProof [EQUIVALENT, 0 ms]
        (46) QDP
        (47) UsableRulesProof [EQUIVALENT, 0 ms]
        (48) QDP
        (49) QReductionProof [EQUIVALENT, 0 ms]
        (50) QDP
        (51) TransformationProof [EQUIVALENT, 0 ms]
        (52) QDP
        (53) TransformationProof [EQUIVALENT, 0 ms]
        (54) QDP
        (55) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (56) YES
    (57) QDP
        (58) UsableRulesProof [EQUIVALENT, 0 ms]
        (59) QDP
        (60) QReductionProof [EQUIVALENT, 0 ms]
        (61) QDP
        (62) NonInfProof [EQUIVALENT, 54 ms]
        (63) QDP
        (64) DependencyGraphProof [EQUIVALENT, 0 ms]
        (65) TRUE


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

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

   lcm(x, y) -> lcmIter(x, y, 0, times(x, y))
   lcmIter(x, y, z, u) -> if(or(ge(0, x), ge(z, u)), x, y, z, u)
   if(true, x, y, z, u) -> z
   if(false, x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
   if2(true, x, y, z, u) -> z
   if2(false, x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(x, y) -> ifTimes(ge(0, x), x, y)
   ifTimes(true, x, y) -> 0
   ifTimes(false, x, y) -> plus(y, times(y, p(x)))
   p(s(x)) -> x
   p(0) -> s(s(0))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   or(true, y) -> true
   or(false, y) -> y
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false
   div(s(x), y, s(z)) -> div(x, y, z)
   a -> b
   a -> c

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   if(true, x, y, z, u) -> z
   if(false, x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
   if2(true, x, y, z, u) -> z
   if2(false, x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(x, y) -> ifTimes(ge(0, x), x, y)
   ifTimes(true, x, y) -> 0
   ifTimes(false, x, y) -> plus(y, times(y, p(x)))
   p(s(x)) -> x
   p(0) -> s(s(0))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   or(true, y) -> true
   or(false, y) -> y
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false
   div(s(x), y, s(z)) -> div(x, y, z)
   lcm(x, y) -> lcmIter(x, y, 0, times(x, y))
   lcmIter(x, y, z, u) -> if(or(ge(0, x), ge(z, u)), x, y, z, u)

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

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

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

   lcm(x, y) -> lcmIter(x, y, 0, times(x, y))
   lcmIter(x, y, z, u) -> if(or(ge(0, x), ge(z, u)), x, y, z, u)
   if(true, x, y, z, u) -> z
   if(false, x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
   if2(true, x, y, z, u) -> z
   if2(false, x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(x, y) -> ifTimes(ge(0, x), x, y)
   ifTimes(true, x, y) -> 0
   ifTimes(false, x, y) -> plus(y, times(y, p(x)))
   p(s(x)) -> x
   p(0) -> s(s(0))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   or(true, y) -> true
   or(false, y) -> y
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false
   div(s(x), y, s(z)) -> div(x, y, z)
   a -> b
   a -> c

The set Q consists of the following terms:

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a


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

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

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

   LCM(x, y) -> LCMITER(x, y, 0, times(x, y))
   LCM(x, y) -> TIMES(x, y)
   LCMITER(x, y, z, u) -> IF(or(ge(0, x), ge(z, u)), x, y, z, u)
   LCMITER(x, y, z, u) -> OR(ge(0, x), ge(z, u))
   LCMITER(x, y, z, u) -> GE(0, x)
   LCMITER(x, y, z, u) -> GE(z, u)
   IF(false, x, y, z, u) -> IF2(divisible(z, y), x, y, z, u)
   IF(false, x, y, z, u) -> DIVISIBLE(z, y)
   IF2(false, x, y, z, u) -> LCMITER(x, y, plus(x, z), u)
   IF2(false, x, y, z, u) -> PLUS(x, z)
   PLUS(s(x), y) -> PLUS(x, y)
   TIMES(x, y) -> IFTIMES(ge(0, x), x, y)
   TIMES(x, y) -> GE(0, x)
   IFTIMES(false, x, y) -> PLUS(y, times(y, p(x)))
   IFTIMES(false, x, y) -> TIMES(y, p(x))
   IFTIMES(false, x, y) -> P(x)
   GE(s(x), s(y)) -> GE(x, y)
   DIVISIBLE(s(x), s(y)) -> DIV(s(x), s(y), s(y))
   DIV(x, y, 0) -> DIVISIBLE(x, y)
   DIV(s(x), y, s(z)) -> DIV(x, y, z)

The TRS R consists of the following rules:

   lcm(x, y) -> lcmIter(x, y, 0, times(x, y))
   lcmIter(x, y, z, u) -> if(or(ge(0, x), ge(z, u)), x, y, z, u)
   if(true, x, y, z, u) -> z
   if(false, x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
   if2(true, x, y, z, u) -> z
   if2(false, x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(x, y) -> ifTimes(ge(0, x), x, y)
   ifTimes(true, x, y) -> 0
   ifTimes(false, x, y) -> plus(y, times(y, p(x)))
   p(s(x)) -> x
   p(0) -> s(s(0))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   or(true, y) -> true
   or(false, y) -> y
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false
   div(s(x), y, s(z)) -> div(x, y, z)
   a -> b
   a -> c

The set Q consists of the following terms:

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 10 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   DIV(s(x), y, s(z)) -> DIV(x, y, z)
   DIV(x, y, 0) -> DIVISIBLE(x, y)
   DIVISIBLE(s(x), s(y)) -> DIV(s(x), s(y), s(y))

The TRS R consists of the following rules:

   lcm(x, y) -> lcmIter(x, y, 0, times(x, y))
   lcmIter(x, y, z, u) -> if(or(ge(0, x), ge(z, u)), x, y, z, u)
   if(true, x, y, z, u) -> z
   if(false, x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
   if2(true, x, y, z, u) -> z
   if2(false, x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(x, y) -> ifTimes(ge(0, x), x, y)
   ifTimes(true, x, y) -> 0
   ifTimes(false, x, y) -> plus(y, times(y, p(x)))
   p(s(x)) -> x
   p(0) -> s(s(0))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   or(true, y) -> true
   or(false, y) -> y
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false
   div(s(x), y, s(z)) -> div(x, y, z)
   a -> b
   a -> c

The set Q consists of the following terms:

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a

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

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

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

   DIV(s(x), y, s(z)) -> DIV(x, y, z)
   DIV(x, y, 0) -> DIVISIBLE(x, y)
   DIVISIBLE(s(x), s(y)) -> DIV(s(x), s(y), s(y))

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

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a

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

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

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a


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

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

   DIV(s(x), y, s(z)) -> DIV(x, y, z)
   DIV(x, y, 0) -> DIVISIBLE(x, y)
   DIVISIBLE(s(x), s(y)) -> DIV(s(x), s(y), s(y))

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

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

From the DPs we obtained the following set of size-change graphs:
*DIV(s(x), y, s(z)) -> DIV(x, y, z)
The graph contains the following edges 1 > 1, 2 >= 2, 3 > 3


*DIV(x, y, 0) -> DIVISIBLE(x, y)
The graph contains the following edges 1 >= 1, 2 >= 2


*DIVISIBLE(s(x), s(y)) -> DIV(s(x), s(y), s(y))
The graph contains the following edges 1 >= 1, 2 >= 2, 2 >= 3


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

(13)
YES

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

(14)
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:

   lcm(x, y) -> lcmIter(x, y, 0, times(x, y))
   lcmIter(x, y, z, u) -> if(or(ge(0, x), ge(z, u)), x, y, z, u)
   if(true, x, y, z, u) -> z
   if(false, x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
   if2(true, x, y, z, u) -> z
   if2(false, x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(x, y) -> ifTimes(ge(0, x), x, y)
   ifTimes(true, x, y) -> 0
   ifTimes(false, x, y) -> plus(y, times(y, p(x)))
   p(s(x)) -> x
   p(0) -> s(s(0))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   or(true, y) -> true
   or(false, y) -> y
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false
   div(s(x), y, s(z)) -> div(x, y, z)
   a -> b
   a -> c

The set Q consists of the following terms:

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a

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

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

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

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

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

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a

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

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

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a


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

(18)
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.
----------------------------------------

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

From the DPs we obtained the following set of size-change graphs:
*GE(s(x), s(y)) -> GE(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(20)
YES

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

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

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

The TRS R consists of the following rules:

   lcm(x, y) -> lcmIter(x, y, 0, times(x, y))
   lcmIter(x, y, z, u) -> if(or(ge(0, x), ge(z, u)), x, y, z, u)
   if(true, x, y, z, u) -> z
   if(false, x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
   if2(true, x, y, z, u) -> z
   if2(false, x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(x, y) -> ifTimes(ge(0, x), x, y)
   ifTimes(true, x, y) -> 0
   ifTimes(false, x, y) -> plus(y, times(y, p(x)))
   p(s(x)) -> x
   p(0) -> s(s(0))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   or(true, y) -> true
   or(false, y) -> y
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false
   div(s(x), y, s(z)) -> div(x, y, z)
   a -> b
   a -> c

The set Q consists of the following terms:

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a

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

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

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

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

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

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a

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

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

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a


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

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

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

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

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

From the DPs we obtained the following set of size-change graphs:
*PLUS(s(x), y) -> PLUS(x, y)
The graph contains the following edges 1 > 1, 2 >= 2


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

(27)
YES

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

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

   IFTIMES(false, x, y) -> TIMES(y, p(x))
   TIMES(x, y) -> IFTIMES(ge(0, x), x, y)

The TRS R consists of the following rules:

   lcm(x, y) -> lcmIter(x, y, 0, times(x, y))
   lcmIter(x, y, z, u) -> if(or(ge(0, x), ge(z, u)), x, y, z, u)
   if(true, x, y, z, u) -> z
   if(false, x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
   if2(true, x, y, z, u) -> z
   if2(false, x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(x, y) -> ifTimes(ge(0, x), x, y)
   ifTimes(true, x, y) -> 0
   ifTimes(false, x, y) -> plus(y, times(y, p(x)))
   p(s(x)) -> x
   p(0) -> s(s(0))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   or(true, y) -> true
   or(false, y) -> y
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false
   div(s(x), y, s(z)) -> div(x, y, z)
   a -> b
   a -> c

The set Q consists of the following terms:

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a

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

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

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

   IFTIMES(false, x, y) -> TIMES(y, p(x))
   TIMES(x, y) -> IFTIMES(ge(0, x), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a

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

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

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a


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

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

   IFTIMES(false, x, y) -> TIMES(y, p(x))
   TIMES(x, y) -> IFTIMES(ge(0, x), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(33) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule TIMES(x, y) -> IFTIMES(ge(0, x), x, y) at position [0] we obtained the following new rules [LPAR04]:

   (TIMES(0, y1) -> IFTIMES(true, 0, y1),TIMES(0, y1) -> IFTIMES(true, 0, y1))
   (TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1),TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1))


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

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

   IFTIMES(false, x, y) -> TIMES(y, p(x))
   TIMES(0, y1) -> IFTIMES(true, 0, y1)
   TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

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

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

   TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1)
   IFTIMES(false, x, y) -> TIMES(y, p(x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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:

   TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1)
   IFTIMES(false, x, y) -> TIMES(y, p(x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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].

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


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

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

   TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1)
   IFTIMES(false, x, y) -> TIMES(y, p(x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   p(s(x0))
   p(0)

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

(41) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IFTIMES(false, x, y) -> TIMES(y, p(x)) we obtained the following new rules [LPAR04]:

   (IFTIMES(false, s(z0), z1) -> TIMES(z1, p(s(z0))),IFTIMES(false, s(z0), z1) -> TIMES(z1, p(s(z0))))


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

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

   TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1)
   IFTIMES(false, s(z0), z1) -> TIMES(z1, p(s(z0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   p(s(x0))
   p(0)

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

(43) 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.
----------------------------------------

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

   TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1)
   IFTIMES(false, s(z0), z1) -> TIMES(z1, p(s(z0)))

The TRS R consists of the following rules:

   p(s(x)) -> x

The set Q consists of the following terms:

   p(s(x0))
   p(0)

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

(45) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule IFTIMES(false, s(z0), z1) -> TIMES(z1, p(s(z0))) at position [1] we obtained the following new rules [LPAR04]:

   (IFTIMES(false, s(z0), z1) -> TIMES(z1, z0),IFTIMES(false, s(z0), z1) -> TIMES(z1, z0))


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

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

   TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1)
   IFTIMES(false, s(z0), z1) -> TIMES(z1, z0)

The TRS R consists of the following rules:

   p(s(x)) -> x

The set Q consists of the following terms:

   p(s(x0))
   p(0)

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

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

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

   TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1)
   IFTIMES(false, s(z0), z1) -> TIMES(z1, z0)

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

   p(s(x0))
   p(0)

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

(49) 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].

   p(s(x0))
   p(0)


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

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

   TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1)
   IFTIMES(false, s(z0), z1) -> TIMES(z1, z0)

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

(51) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule IFTIMES(false, s(z0), z1) -> TIMES(z1, z0) we obtained the following new rules [LPAR04]:

   (IFTIMES(false, s(x0), s(y_0)) -> TIMES(s(y_0), x0),IFTIMES(false, s(x0), s(y_0)) -> TIMES(s(y_0), x0))


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

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

   TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1)
   IFTIMES(false, s(x0), s(y_0)) -> TIMES(s(y_0), x0)

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

(53) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule TIMES(s(x0), y1) -> IFTIMES(false, s(x0), y1) we obtained the following new rules [LPAR04]:

   (TIMES(s(x0), s(y_1)) -> IFTIMES(false, s(x0), s(y_1)),TIMES(s(x0), s(y_1)) -> IFTIMES(false, s(x0), s(y_1)))


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

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

   IFTIMES(false, s(x0), s(y_0)) -> TIMES(s(y_0), x0)
   TIMES(s(x0), s(y_1)) -> IFTIMES(false, s(x0), s(y_1))

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

(55) 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:
*TIMES(s(x0), s(y_1)) -> IFTIMES(false, s(x0), s(y_1))
The graph contains the following edges 1 >= 2, 2 >= 3


*IFTIMES(false, s(x0), s(y_0)) -> TIMES(s(y_0), x0)
The graph contains the following edges 3 >= 1, 2 > 2


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

(56)
YES

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

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

   IF(false, x, y, z, u) -> IF2(divisible(z, y), x, y, z, u)
   IF2(false, x, y, z, u) -> LCMITER(x, y, plus(x, z), u)
   LCMITER(x, y, z, u) -> IF(or(ge(0, x), ge(z, u)), x, y, z, u)

The TRS R consists of the following rules:

   lcm(x, y) -> lcmIter(x, y, 0, times(x, y))
   lcmIter(x, y, z, u) -> if(or(ge(0, x), ge(z, u)), x, y, z, u)
   if(true, x, y, z, u) -> z
   if(false, x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
   if2(true, x, y, z, u) -> z
   if2(false, x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(x, y) -> ifTimes(ge(0, x), x, y)
   ifTimes(true, x, y) -> 0
   ifTimes(false, x, y) -> plus(y, times(y, p(x)))
   p(s(x)) -> x
   p(0) -> s(s(0))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   or(true, y) -> true
   or(false, y) -> y
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false
   div(s(x), y, s(z)) -> div(x, y, z)
   a -> b
   a -> c

The set Q consists of the following terms:

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a

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:

   IF(false, x, y, z, u) -> IF2(divisible(z, y), x, y, z, u)
   IF2(false, x, y, z, u) -> LCMITER(x, y, plus(x, z), u)
   LCMITER(x, y, z, u) -> IF(or(ge(0, x), ge(z, u)), x, y, z, u)

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)
   or(true, y) -> true
   or(false, y) -> y
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(s(x), y, s(z)) -> div(x, y, z)
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false

The set Q consists of the following terms:

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   plus(0, x0)
   plus(s(x0), x1)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))
   a

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].

   lcm(x0, x1)
   lcmIter(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   if2(true, x0, x1, x2, x3)
   if2(false, x0, x1, x2, x3)
   times(x0, x1)
   ifTimes(true, x0, x1)
   ifTimes(false, x0, x1)
   p(s(x0))
   p(0)
   a


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

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

   IF(false, x, y, z, u) -> IF2(divisible(z, y), x, y, z, u)
   IF2(false, x, y, z, u) -> LCMITER(x, y, plus(x, z), u)
   LCMITER(x, y, z, u) -> IF(or(ge(0, x), ge(z, u)), x, y, z, u)

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)
   or(true, y) -> true
   or(false, y) -> y
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(s(x), y, s(z)) -> div(x, y, z)
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false

The set Q consists of the following terms:

   plus(0, x0)
   plus(s(x0), x1)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))

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

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

Note that final constraints are written in bold face.



For Pair IF(false, x, y, z, u) -> IF2(divisible(z, y), x, y, z, u) the following chains were created:
*We consider the chain LCMITER(x8, x9, x10, x11) -> IF(or(ge(0, x8), ge(x10, x11)), x8, x9, x10, x11), IF(false, x12, x13, x14, x15) -> IF2(divisible(x14, x13), x12, x13, x14, x15) which results in the following constraint:

(1)    (IF(or(ge(0, x8), ge(x10, x11)), x8, x9, x10, x11)=IF(false, x12, x13, x14, x15)  ==>  IF(false, x12, x13, x14, x15)_>=_IF2(divisible(x14, x13), x12, x13, x14, x15))



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

(2)    (0=x50 & ge(x50, x8)=x48 & ge(x10, x11)=x49 & or(x48, x49)=false  ==>  IF(false, x8, x9, x10, x11)_>=_IF2(divisible(x10, x9), x8, x9, x10, x11))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on or(x48, x49)=false which results in the following new constraint:

(3)    (x52=false & 0=x50 & ge(x50, x8)=false & ge(x10, x11)=x52  ==>  IF(false, x8, x9, x10, x11)_>=_IF2(divisible(x10, x9), x8, x9, x10, x11))



We simplified constraint (3) using rule (III) which results in the following new constraint:

(4)    (0=x50 & ge(x50, x8)=false & ge(x10, x11)=false  ==>  IF(false, x8, x9, x10, x11)_>=_IF2(divisible(x10, x9), x8, x9, x10, x11))



We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on ge(x50, x8)=false which results in the following new constraints:

(5)    (false=false & 0=0 & ge(x10, x11)=false  ==>  IF(false, s(x54), x9, x10, x11)_>=_IF2(divisible(x10, x9), s(x54), x9, x10, x11))

(6)    (ge(x56, x55)=false & 0=s(x56) & ge(x10, x11)=false & (\/x57,x58,x59:ge(x56, x55)=false & 0=x56 & ge(x57, x58)=false  ==>  IF(false, x55, x59, x57, x58)_>=_IF2(divisible(x57, x59), x55, x59, x57, x58))  ==>  IF(false, s(x55), x9, x10, x11)_>=_IF2(divisible(x10, x9), s(x55), x9, x10, x11))



We simplified constraint (5) using rules (I), (II) which results in the following new constraint:

(7)    (ge(x10, x11)=false  ==>  IF(false, s(x54), x9, x10, x11)_>=_IF2(divisible(x10, x9), s(x54), x9, x10, x11))



We solved constraint (6) using rules (I), (II).We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on ge(x10, x11)=false which results in the following new constraints:

(8)    (false=false  ==>  IF(false, s(x54), x9, 0, s(x61))_>=_IF2(divisible(0, x9), s(x54), x9, 0, s(x61)))

(9)    (ge(x63, x62)=false & (\/x64,x65:ge(x63, x62)=false  ==>  IF(false, s(x64), x65, x63, x62)_>=_IF2(divisible(x63, x65), s(x64), x65, x63, x62))  ==>  IF(false, s(x54), x9, s(x63), s(x62))_>=_IF2(divisible(s(x63), x9), s(x54), x9, s(x63), s(x62)))



We simplified constraint (8) using rules (I), (II) which results in the following new constraint:

(10)    (IF(false, s(x54), x9, 0, s(x61))_>=_IF2(divisible(0, x9), s(x54), x9, 0, s(x61)))



We simplified constraint (9) using rule (VI) where we applied the induction hypothesis (\/x64,x65:ge(x63, x62)=false  ==>  IF(false, s(x64), x65, x63, x62)_>=_IF2(divisible(x63, x65), s(x64), x65, x63, x62)) with sigma = [x64 / x54, x65 / x9] which results in the following new constraint:

(11)    (IF(false, s(x54), x9, x63, x62)_>=_IF2(divisible(x63, x9), s(x54), x9, x63, x62)  ==>  IF(false, s(x54), x9, s(x63), s(x62))_>=_IF2(divisible(s(x63), x9), s(x54), x9, s(x63), s(x62)))








For Pair IF2(false, x, y, z, u) -> LCMITER(x, y, plus(x, z), u) the following chains were created:
*We consider the chain IF(false, x16, x17, x18, x19) -> IF2(divisible(x18, x17), x16, x17, x18, x19), IF2(false, x20, x21, x22, x23) -> LCMITER(x20, x21, plus(x20, x22), x23) which results in the following constraint:

(1)    (IF2(divisible(x18, x17), x16, x17, x18, x19)=IF2(false, x20, x21, x22, x23)  ==>  IF2(false, x20, x21, x22, x23)_>=_LCMITER(x20, x21, plus(x20, x22), x23))



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

(2)    (divisible(x18, x17)=false  ==>  IF2(false, x16, x17, x18, x19)_>=_LCMITER(x16, x17, plus(x16, x18), x19))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on divisible(x18, x17)=false which results in the following new constraint:

(3)    (div(s(x68), s(x67), s(x67))=false  ==>  IF2(false, x16, s(x67), s(x68), x19)_>=_LCMITER(x16, s(x67), plus(x16, s(x68)), x19))



We simplified constraint (3) using rule (VII) which results in the following new constraint:

(4)    (s(x68)=x69 & s(x67)=x70 & s(x67)=x71 & div(x69, x70, x71)=false  ==>  IF2(false, x16, s(x67), s(x68), x19)_>=_LCMITER(x16, s(x67), plus(x16, s(x68)), x19))



We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on div(x69, x70, x71)=false which results in the following new constraints:

(5)    (div(x74, x73, x72)=false & s(x68)=s(x74) & s(x67)=x73 & s(x67)=s(x72) & (\/x75,x76,x77,x78:div(x74, x73, x72)=false & s(x75)=x74 & s(x76)=x73 & s(x76)=x72  ==>  IF2(false, x77, s(x76), s(x75), x78)_>=_LCMITER(x77, s(x76), plus(x77, s(x75)), x78))  ==>  IF2(false, x16, s(x67), s(x68), x19)_>=_LCMITER(x16, s(x67), plus(x16, s(x68)), x19))

(6)    (divisible(x80, x79)=false & s(x68)=x80 & s(x67)=x79 & s(x67)=0  ==>  IF2(false, x16, s(x67), s(x68), x19)_>=_LCMITER(x16, s(x67), plus(x16, s(x68)), x19))

(7)    (false=false & s(x68)=0 & s(x67)=x82 & s(x67)=s(x81)  ==>  IF2(false, x16, s(x67), s(x68), x19)_>=_LCMITER(x16, s(x67), plus(x16, s(x68)), x19))



We simplified constraint (5) using rules (I), (II), (III), (IV) which results in the following new constraint:

(8)    (IF2(false, x16, s(x72), s(x74), x19)_>=_LCMITER(x16, s(x72), plus(x16, s(x74)), x19))



We solved constraint (6) using rules (I), (II).We solved constraint (7) using rules (I), (II).




For Pair LCMITER(x, y, z, u) -> IF(or(ge(0, x), ge(z, u)), x, y, z, u) the following chains were created:
*We consider the chain IF2(false, x36, x37, x38, x39) -> LCMITER(x36, x37, plus(x36, x38), x39), LCMITER(x40, x41, x42, x43) -> IF(or(ge(0, x40), ge(x42, x43)), x40, x41, x42, x43) which results in the following constraint:

(1)    (LCMITER(x36, x37, plus(x36, x38), x39)=LCMITER(x40, x41, x42, x43)  ==>  LCMITER(x40, x41, x42, x43)_>=_IF(or(ge(0, x40), ge(x42, x43)), x40, x41, x42, x43))



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

(2)    (LCMITER(x36, x37, x42, x39)_>=_IF(or(ge(0, x36), ge(x42, x39)), x36, x37, x42, x39))








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

*IF(false, x, y, z, u) -> IF2(divisible(z, y), x, y, z, u)

*(IF(false, s(x54), x9, 0, s(x61))_>=_IF2(divisible(0, x9), s(x54), x9, 0, s(x61)))


*(IF(false, s(x54), x9, x63, x62)_>=_IF2(divisible(x63, x9), s(x54), x9, x63, x62)  ==>  IF(false, s(x54), x9, s(x63), s(x62))_>=_IF2(divisible(s(x63), x9), s(x54), x9, s(x63), s(x62)))




*IF2(false, x, y, z, u) -> LCMITER(x, y, plus(x, z), u)

*(IF2(false, x16, s(x72), s(x74), x19)_>=_LCMITER(x16, s(x72), plus(x16, s(x74)), x19))




*LCMITER(x, y, z, u) -> IF(or(ge(0, x), ge(z, u)), x, y, z, u)

*(LCMITER(x36, x37, x42, x39)_>=_IF(or(ge(0, x36), ge(x42, x39)), x36, x37, x42, x39))








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

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

Polynomial interpretation [NONINF]:

   POL(0) = 0
   POL(IF(x_1, x_2, x_3, x_4, x_5)) = -1 - x_1 + x_3 - x_4 + x_5
   POL(IF2(x_1, x_2, x_3, x_4, x_5)) = -1 - x_1 - x_2 + x_3 - x_4 + x_5
   POL(LCMITER(x_1, x_2, x_3, x_4)) = -1 + x_2 - x_3 + x_4
   POL(c) = -1
   POL(div(x_1, x_2, x_3)) = 0
   POL(divisible(x_1, x_2)) = 0
   POL(false) = 0
   POL(ge(x_1, x_2)) = 0
   POL(or(x_1, x_2)) = 0
   POL(plus(x_1, x_2)) = x_1 + x_2
   POL(s(x_1)) = 1 + x_1
   POL(true) = 0


The following pairs  are in P_>:
   IF(false, x, y, z, u) -> IF2(divisible(z, y), x, y, z, u)
The following pairs are in P_bound:
   IF(false, x, y, z, u) -> IF2(divisible(z, y), x, y, z, u)
The following rules are usable:
   true -> divisible(0, s(y))
   div(s(x), s(y), s(y)) -> divisible(s(x), s(y))
   div(x, y, z) -> div(s(x), y, s(z))
   divisible(x, y) -> div(x, y, 0)
   y -> plus(0, y)
   s(plus(x, y)) -> plus(s(x), y)
   true -> or(true, y)
   y -> or(false, y)
   false -> div(0, y, s(z))

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

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

   IF2(false, x, y, z, u) -> LCMITER(x, y, plus(x, z), u)
   LCMITER(x, y, z, u) -> IF(or(ge(0, x), ge(z, u)), x, y, z, u)

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)
   or(true, y) -> true
   or(false, y) -> y
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   divisible(0, s(y)) -> true
   divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
   div(s(x), y, s(z)) -> div(x, y, z)
   div(x, y, 0) -> divisible(x, y)
   div(0, y, s(z)) -> false

The set Q consists of the following terms:

   plus(0, x0)
   plus(s(x0), x1)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   or(true, x0)
   or(false, x0)
   divisible(0, s(x0))
   divisible(s(x0), s(x1))
   div(x0, x1, 0)
   div(0, x0, s(x1))
   div(s(x0), x1, s(x2))

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

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

(65)
TRUE