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)