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 (121ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (6ms), PolynomialLinearRange4iUR (954ms), DependencyGraph (4ms), PolynomialLinearRange8NegiUR (5369ms), DependencyGraph (3ms), ReductionPairSAT (1636ms), DependencyGraph (4ms), SizeChangePrinciple (398ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (4ms), Propagation (1ms)].
 | – Problem 4 was processed with processor SubtermCriterion (0ms).
 | – Problem 5 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

plus#(id(x), s(y))plus#(x, if(gt(s(y), y), y, s(y)))plus#(s(x), s(y))plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
plus#(s(x), x)plus#(if(gt(x, x), id(x), id(x)), s(x))

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
double(0)0double(s(x))s(s(double(x)))
plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y)yplus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
gt(s(x), zero)truegt(zero, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: not, id, plus, 0, minus, s, if, false, true, gt, zero, double


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

plus#(s(x), x)id#(x)plus#(id(x), s(y))plus#(x, if(gt(s(y), y), y, s(y)))
plus#(id(x), s(y))gt#(s(y), y)plus#(s(x), s(y))gt#(x, y)
plus#(s(x), s(y))if#(gt(x, y), x, y)plus#(s(x), x)plus#(if(gt(x, x), id(x), id(x)), s(x))
plus#(s(x), x)gt#(x, x)plus#(id(x), s(y))if#(gt(s(y), y), y, s(y))
plus#(s(x), s(y))id#(x)plus#(s(x), x)if#(gt(x, x), id(x), id(x))
plus#(s(x), s(y))plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))gt#(s(x), s(y))gt#(x, y)
plus#(s(x), s(y))if#(not(gt(x, y)), id(x), id(y))minus#(s(x), s(y))minus#(x, y)
double#(s(x))double#(x)not#(x)if#(x, false, true)
plus#(s(x), s(y))id#(y)plus#(s(x), s(y))not#(gt(x, y))

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
double(0)0double(s(x))s(s(double(x)))
plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y)yplus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
gt(s(x), zero)truegt(zero, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: id, not, plus, minus, 0, s, if, true, false, gt, double, zero

Strategy


The following SCCs where found

plus#(id(x), s(y)) → plus#(x, if(gt(s(y), y), y, s(y)))plus#(s(x), s(y)) → plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
plus#(s(x), x) → plus#(if(gt(x, x), id(x), id(x)), s(x))

gt#(s(x), s(y)) → gt#(x, y)

minus#(s(x), s(y)) → minus#(x, y)

double#(s(x)) → double#(x)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

double#(s(x))double#(x)

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
double(0)0double(s(x))s(s(double(x)))
plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y)yplus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
gt(s(x), zero)truegt(zero, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: id, not, plus, minus, 0, s, if, true, false, gt, double, zero

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

double#(s(x))double#(x)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

minus#(s(x), s(y))minus#(x, y)

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
double(0)0double(s(x))s(s(double(x)))
plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y)yplus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
gt(s(x), zero)truegt(zero, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: id, not, plus, minus, 0, s, if, true, false, gt, double, zero

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

minus#(s(x), s(y))minus#(x, y)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

gt#(s(x), s(y))gt#(x, y)

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
double(0)0double(s(x))s(s(double(x)))
plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y)yplus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
gt(s(x), zero)truegt(zero, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: id, not, plus, minus, 0, s, if, true, false, gt, double, zero

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

gt#(s(x), s(y))gt#(x, y)