TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60042 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (64ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (252ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (852ms), DependencyGraph (3ms), ReductionPairSAT (timeout)].
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (0ms).
 | – Problem 5 was processed with processor SubtermCriterion (1ms).
 | – Problem 6 was processed with processor SubtermCriterion (0ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

help#(false, c, x, y, z)towerIter#(s(c), x, y, exp(y, z))towerIter#(c, x, y, z)help#(ge(c, x), c, x, y, z)

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
ge(x, 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)tower(x, y)towerIter(0, x, y, s(0))
towerIter(c, x, y, z)help(ge(c, x), c, x, y, z)help(true, c, x, y, z)z
help(false, c, x, y, z)towerIter(s(c), x, y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: exp, plus, help, tower, 0, s, times, towerIter, false, true, ge


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

times#(s(x), y)plus#(y, times(x, y))towerIter#(c, x, y, z)ge#(c, x)
exp#(x, s(y))exp#(x, y)help#(false, c, x, y, z)towerIter#(s(c), x, y, exp(y, z))
times#(s(x), y)times#(x, y)tower#(x, y)towerIter#(0, x, y, s(0))
plus#(s(x), y)plus#(x, y)ge#(s(x), s(y))ge#(x, y)
exp#(x, s(y))times#(x, exp(x, y))towerIter#(c, x, y, z)help#(ge(c, x), c, x, y, z)
help#(false, c, x, y, z)exp#(y, z)

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
ge(x, 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)tower(x, y)towerIter(0, x, y, s(0))
towerIter(c, x, y, z)help(ge(c, x), c, x, y, z)help(true, c, x, y, z)z
help(false, c, x, y, z)towerIter(s(c), x, y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, exp, tower, help, 0, s, times, true, false, towerIter, ge

Strategy


The following SCCs where found

exp#(x, s(y)) → exp#(x, y)

help#(false, c, x, y, z) → towerIter#(s(c), x, y, exp(y, z))towerIter#(c, x, y, z) → help#(ge(c, x), c, x, y, z)

times#(s(x), y) → times#(x, y)

plus#(s(x), y) → plus#(x, y)

ge#(s(x), s(y)) → ge#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

exp#(x, s(y))exp#(x, y)

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
ge(x, 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)tower(x, y)towerIter(0, x, y, s(0))
towerIter(c, x, y, z)help(ge(c, x), c, x, y, z)help(true, c, x, y, z)z
help(false, c, x, y, z)towerIter(s(c), x, y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, exp, tower, help, 0, s, times, true, false, towerIter, ge

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

exp#(x, s(y))exp#(x, y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
ge(x, 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)tower(x, y)towerIter(0, x, y, s(0))
towerIter(c, x, y, z)help(ge(c, x), c, x, y, z)help(true, c, x, y, z)z
help(false, c, x, y, z)towerIter(s(c), x, y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, exp, tower, help, 0, s, times, true, false, towerIter, ge

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
ge(x, 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)tower(x, y)towerIter(0, x, y, s(0))
towerIter(c, x, y, z)help(ge(c, x), c, x, y, z)help(true, c, x, y, z)z
help(false, c, x, y, z)towerIter(s(c), x, y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, exp, tower, help, 0, s, times, true, false, towerIter, ge

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

times#(s(x), y)times#(x, y)

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
exp(x, 0)s(0)exp(x, s(y))times(x, exp(x, y))
ge(x, 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)tower(x, y)towerIter(0, x, y, s(0))
towerIter(c, x, y, z)help(ge(c, x), c, x, y, z)help(true, c, x, y, z)z
help(false, c, x, y, z)towerIter(s(c), x, y, exp(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, exp, tower, help, 0, s, times, true, false, towerIter, ge

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

times#(s(x), y)times#(x, y)