(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) 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.
(9) 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.
(10) 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.
(11) 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.