Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 AAECC Innermost (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 DependencyGraphProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE
  ↳22 QDP
  ↳23 QDP

(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 transformation)

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 {c, a, b}

(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 transformation)

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 transformation)

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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DIV(s(x), y, s(z)) → DIV(x, y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2, x3)  =  DIV(x1, x2)
s(x1)  =  s(x1)
0  =  0
DIVISIBLE(x1, x2)  =  DIVISIBLE(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
s1 > [DIV2, DIVISIBLE2]
0 > [DIV2, DIVISIBLE2]

Status:
DIV2: [1,2]
s1: [1]
DIVISIBLE2: [1,2]
0: []


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

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.

(10) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(11) TRUE

(12) 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.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


GE(s(x), s(y)) → GE(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GE(x1, x2)  =  GE(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
s1: [1]
GE1: [1]


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
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) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) 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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PLUS(s(x), y) → PLUS(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
[PLUS2, s1]

Status:
PLUS2: [2,1]
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

Q DP problem:
P is empty.
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.

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE

(22) 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.

(23) 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.