TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60000 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (206ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (8ms), PolynomialLinearRange4iUR (827ms), DependencyGraph (6ms), PolynomialLinearRange8NegiUR (10156ms), DependencyGraph (5ms), ReductionPairSAT (timeout)].
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 7 was processed with processor DependencyGraph (5ms).
 | – Problem 5 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (4ms), PolynomialLinearRange4iUR (105ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (449ms), DependencyGraph (1ms)].
 | – Problem 6 was processed with processor SubtermCriterion (2ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

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)

Rewrite 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)zif(false, x, y, z, u)if2(divisible(z, y), x, y, z, u)
if2(true, x, y, z, u)zif2(false, x, y, z, u)lcmIter(x, y, plus(x, z), u)
plus(0, y)yplus(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))falsege(s(x), s(y))ge(x, y)
or(true, y)trueor(false, y)y
divisible(0, s(y))truedivisible(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)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, b, c, or, a, div, true, ge, if2, lcmIter, 0, lcm, divisible, s, times, if, p, false




Open Dependency Pair Problem 5

Dependency Pairs

ifTimes#(false, x, y)times#(y, p(x))times#(x, y)ifTimes#(ge(0, x), x, y)

Rewrite 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)zif(false, x, y, z, u)if2(divisible(z, y), x, y, z, u)
if2(true, x, y, z, u)zif2(false, x, y, z, u)lcmIter(x, y, plus(x, z), u)
plus(0, y)yplus(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))falsege(s(x), s(y))ge(x, y)
or(true, y)trueor(false, y)y
divisible(0, s(y))truedivisible(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)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, b, c, or, a, div, true, ge, if2, lcmIter, 0, lcm, divisible, s, times, if, p, false


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

lcmIter#(x, y, z, u)or#(ge(0, x), ge(z, u))lcmIter#(x, y, z, u)ge#(z, u)
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)
ifTimes#(false, x, y)times#(y, p(x))lcmIter#(x, y, z, u)if#(or(ge(0, x), ge(z, u)), x, y, z, u)
if#(false, x, y, z, u)divisible#(z, y)div#(x, y, 0)divisible#(x, y)
times#(x, y)ifTimes#(ge(0, x), x, y)divisible#(s(x), s(y))div#(s(x), s(y), s(y))
ifTimes#(false, x, y)plus#(y, times(y, p(x)))times#(x, y)ge#(0, x)
lcm#(x, y)times#(x, y)div#(s(x), y, s(z))div#(x, y, z)
plus#(s(x), y)plus#(x, y)lcm#(x, y)lcmIter#(x, y, 0, times(x, y))
lcmIter#(x, y, z, u)ge#(0, x)ifTimes#(false, x, y)p#(x)
ge#(s(x), s(y))ge#(x, y)if2#(false, x, y, z, u)plus#(x, z)

Rewrite 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)zif(false, x, y, z, u)if2(divisible(z, y), x, y, z, u)
if2(true, x, y, z, u)zif2(false, x, y, z, u)lcmIter(x, y, plus(x, z), u)
plus(0, y)yplus(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))falsege(s(x), s(y))ge(x, y)
or(true, y)trueor(false, y)y
divisible(0, s(y))truedivisible(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)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, b, c, or, a, div, true, ge, if2, lcmIter, 0, lcm, s, divisible, times, if, p, false

Strategy


The following SCCs where found

ifTimes#(false, x, y) → times#(y, p(x))times#(x, y) → ifTimes#(ge(0, x), x, y)

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))

plus#(s(x), y) → plus#(x, y)

ge#(s(x), s(y)) → ge#(x, y)

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)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

plus#(s(x), y)plus#(x, y)

Rewrite 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)zif(false, x, y, z, u)if2(divisible(z, y), x, y, z, u)
if2(true, x, y, z, u)zif2(false, x, y, z, u)lcmIter(x, y, plus(x, z), u)
plus(0, y)yplus(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))falsege(s(x), s(y))ge(x, y)
or(true, y)trueor(false, y)y
divisible(0, s(y))truedivisible(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)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, b, c, or, a, div, true, ge, if2, lcmIter, 0, lcm, s, divisible, times, if, p, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

plus#(s(x), y)plus#(x, y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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))

Rewrite 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)zif(false, x, y, z, u)if2(divisible(z, y), x, y, z, u)
if2(true, x, y, z, u)zif2(false, x, y, z, u)lcmIter(x, y, plus(x, z), u)
plus(0, y)yplus(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))falsege(s(x), s(y))ge(x, y)
or(true, y)trueor(false, y)y
divisible(0, s(y))truedivisible(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)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, b, c, or, a, div, true, ge, if2, lcmIter, 0, lcm, s, divisible, times, if, p, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

div#(s(x), y, s(z))div#(x, y, z)

Problem 7: DependencyGraph



Dependency Pair Problem

Dependency Pairs

div#(x, y, 0)divisible#(x, y)divisible#(s(x), s(y))div#(s(x), s(y), s(y))

Rewrite 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)zif(false, x, y, z, u)if2(divisible(z, y), x, y, z, u)
if2(true, x, y, z, u)zif2(false, x, y, z, u)lcmIter(x, y, plus(x, z), u)
plus(0, y)yplus(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))falsege(s(x), s(y))ge(x, y)
or(true, y)trueor(false, y)y
divisible(0, s(y))truedivisible(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)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, b, c, or, a, div, true, ge, if2, lcmIter, 0, lcm, divisible, s, times, if, p, false

Strategy


There are no SCCs!

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

ge#(s(x), s(y))ge#(x, y)

Rewrite 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)zif(false, x, y, z, u)if2(divisible(z, y), x, y, z, u)
if2(true, x, y, z, u)zif2(false, x, y, z, u)lcmIter(x, y, plus(x, z), u)
plus(0, y)yplus(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))falsege(s(x), s(y))ge(x, y)
or(true, y)trueor(false, y)y
divisible(0, s(y))truedivisible(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)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, b, c, or, a, div, true, ge, if2, lcmIter, 0, lcm, s, divisible, times, if, p, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

ge#(s(x), s(y))ge#(x, y)