TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60095 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (96ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (3ms), PolynomialLinearRange4iUR (232ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (571ms), DependencyGraph (3ms)].
 | – Problem 3 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (380ms), DependencyGraph (6ms), PolynomialLinearRange8NegiUR (1245ms), DependencyGraph (2ms)].
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 | – Problem 5 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

timesIter#(x, y, z, u)ifTimes#(ge(u, x), x, y, z, u)ifTimes#(false, x, y, z, u)timesIter#(x, y, plus(y, z), s(u))

Rewrite Rules

prod(xs)prodIter(xs, s(0))prodIter(xs, x)ifProd(isempty(xs), xs, x)
ifProd(true, xs, x)xifProd(false, xs, x)prodIter(tail(xs), times(x, head(xs)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
times(x, y)timesIter(x, y, 0, 0)timesIter(x, y, z, u)ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u)zifTimes(false, x, y, z, u)timesIter(x, y, plus(y, z), s(u))
isempty(nil)trueisempty(cons(x, xs))false
head(nil)errorhead(cons(x, xs))x
tail(nil)niltail(cons(x, xs))xs
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, ifProd, b, error, c, a, prodIter, true, ge, tail, timesIter, 0, s, isempty, times, false, head, cons, nil, prod




Open Dependency Pair Problem 3

Dependency Pairs

ifProd#(false, xs, x)prodIter#(tail(xs), times(x, head(xs)))prodIter#(xs, x)ifProd#(isempty(xs), xs, x)

Rewrite Rules

prod(xs)prodIter(xs, s(0))prodIter(xs, x)ifProd(isempty(xs), xs, x)
ifProd(true, xs, x)xifProd(false, xs, x)prodIter(tail(xs), times(x, head(xs)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
times(x, y)timesIter(x, y, 0, 0)timesIter(x, y, z, u)ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u)zifTimes(false, x, y, z, u)timesIter(x, y, plus(y, z), s(u))
isempty(nil)trueisempty(cons(x, xs))false
head(nil)errorhead(cons(x, xs))x
tail(nil)niltail(cons(x, xs))xs
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, ifProd, b, error, c, a, prodIter, true, ge, tail, timesIter, 0, s, isempty, times, false, head, cons, nil, prod


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

timesIter#(x, y, z, u)ifTimes#(ge(u, x), x, y, z, u)prodIter#(xs, x)ifProd#(isempty(xs), xs, x)
ifProd#(false, xs, x)head#(xs)ifProd#(false, xs, x)tail#(xs)
times#(x, y)timesIter#(x, y, 0, 0)ifProd#(false, xs, x)prodIter#(tail(xs), times(x, head(xs)))
ifProd#(false, xs, x)times#(x, head(xs))plus#(s(x), y)plus#(x, y)
ifTimes#(false, x, y, z, u)timesIter#(x, y, plus(y, z), s(u))prod#(xs)prodIter#(xs, s(0))
ge#(s(x), s(y))ge#(x, y)ifTimes#(false, x, y, z, u)plus#(y, z)
prodIter#(xs, x)isempty#(xs)timesIter#(x, y, z, u)ge#(u, x)

Rewrite Rules

prod(xs)prodIter(xs, s(0))prodIter(xs, x)ifProd(isempty(xs), xs, x)
ifProd(true, xs, x)xifProd(false, xs, x)prodIter(tail(xs), times(x, head(xs)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
times(x, y)timesIter(x, y, 0, 0)timesIter(x, y, z, u)ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u)zifTimes(false, x, y, z, u)timesIter(x, y, plus(y, z), s(u))
isempty(nil)trueisempty(cons(x, xs))false
head(nil)errorhead(cons(x, xs))x
tail(nil)niltail(cons(x, xs))xs
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, ifProd, b, error, c, a, prodIter, true, ge, tail, timesIter, 0, s, times, isempty, false, head, prod, nil, cons

Strategy


The following SCCs where found

plus#(s(x), y) → plus#(x, y)

ge#(s(x), s(y)) → ge#(x, y)

ifProd#(false, xs, x) → prodIter#(tail(xs), times(x, head(xs)))prodIter#(xs, x) → ifProd#(isempty(xs), xs, x)

timesIter#(x, y, z, u) → ifTimes#(ge(u, x), x, y, z, u)ifTimes#(false, x, y, z, u) → timesIter#(x, y, plus(y, z), s(u))

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

prod(xs)prodIter(xs, s(0))prodIter(xs, x)ifProd(isempty(xs), xs, x)
ifProd(true, xs, x)xifProd(false, xs, x)prodIter(tail(xs), times(x, head(xs)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
times(x, y)timesIter(x, y, 0, 0)timesIter(x, y, z, u)ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u)zifTimes(false, x, y, z, u)timesIter(x, y, plus(y, z), s(u))
isempty(nil)trueisempty(cons(x, xs))false
head(nil)errorhead(cons(x, xs))x
tail(nil)niltail(cons(x, xs))xs
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, ifProd, b, error, c, a, prodIter, true, ge, tail, timesIter, 0, s, times, isempty, false, head, prod, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

prod(xs)prodIter(xs, s(0))prodIter(xs, x)ifProd(isempty(xs), xs, x)
ifProd(true, xs, x)xifProd(false, xs, x)prodIter(tail(xs), times(x, head(xs)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
times(x, y)timesIter(x, y, 0, 0)timesIter(x, y, z, u)ifTimes(ge(u, x), x, y, z, u)
ifTimes(true, x, y, z, u)zifTimes(false, x, y, z, u)timesIter(x, y, plus(y, z), s(u))
isempty(nil)trueisempty(cons(x, xs))false
head(nil)errorhead(cons(x, xs))x
tail(nil)niltail(cons(x, xs))xs
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)ab
ac

Original Signature

Termination of terms over the following signature is verified: ifTimes, plus, ifProd, b, error, c, a, prodIter, true, ge, tail, timesIter, 0, s, times, isempty, false, head, prod, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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