TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60062 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (88ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (463ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (3421ms), 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).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

if#(first, x, y, z)if#(le(s(x), y, s(z)), s(x), y, s(z))if#(second, x, y, z)if#(le(s(x), s(y), z), s(x), s(y), z)

Rewrite Rules

le(0, y, z)greater(y, z)le(s(x), 0, z)false
le(s(x), s(y), 0)falsele(s(x), s(y), s(z))le(x, y, z)
greater(x, 0)firstgreater(0, s(y))second
greater(s(x), s(y))greater(x, y)double(0)0
double(s(x))s(s(double(x)))triple(x)if(le(x, x, double(x)), x, 0, 0)
if(false, x, y, z)trueif(first, x, y, z)if(le(s(x), y, s(z)), s(x), y, s(z))
if(second, x, y, z)if(le(s(x), s(y), z), s(x), s(y), z)

Original Signature

Termination of terms over the following signature is verified: 0, greater, second, s, le, if, triple, true, false, double, first


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

triple#(x)le#(x, x, double(x))le#(s(x), s(y), s(z))le#(x, y, z)
triple#(x)double#(x)if#(first, x, y, z)if#(le(s(x), y, s(z)), s(x), y, s(z))
if#(second, x, y, z)le#(s(x), s(y), z)double#(s(x))double#(x)
le#(0, y, z)greater#(y, z)greater#(s(x), s(y))greater#(x, y)
triple#(x)if#(le(x, x, double(x)), x, 0, 0)if#(first, x, y, z)le#(s(x), y, s(z))
if#(second, x, y, z)if#(le(s(x), s(y), z), s(x), s(y), z)

Rewrite Rules

le(0, y, z)greater(y, z)le(s(x), 0, z)false
le(s(x), s(y), 0)falsele(s(x), s(y), s(z))le(x, y, z)
greater(x, 0)firstgreater(0, s(y))second
greater(s(x), s(y))greater(x, y)double(0)0
double(s(x))s(s(double(x)))triple(x)if(le(x, x, double(x)), x, 0, 0)
if(false, x, y, z)trueif(first, x, y, z)if(le(s(x), y, s(z)), s(x), y, s(z))
if(second, x, y, z)if(le(s(x), s(y), z), s(x), s(y), z)

Original Signature

Termination of terms over the following signature is verified: greater, 0, le, s, second, if, triple, false, true, double, first

Strategy


The following SCCs where found

le#(s(x), s(y), s(z)) → le#(x, y, z)

if#(first, x, y, z) → if#(le(s(x), y, s(z)), s(x), y, s(z))if#(second, x, y, z) → if#(le(s(x), s(y), z), s(x), s(y), z)

double#(s(x)) → double#(x)

greater#(s(x), s(y)) → greater#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

greater#(s(x), s(y))greater#(x, y)

Rewrite Rules

le(0, y, z)greater(y, z)le(s(x), 0, z)false
le(s(x), s(y), 0)falsele(s(x), s(y), s(z))le(x, y, z)
greater(x, 0)firstgreater(0, s(y))second
greater(s(x), s(y))greater(x, y)double(0)0
double(s(x))s(s(double(x)))triple(x)if(le(x, x, double(x)), x, 0, 0)
if(false, x, y, z)trueif(first, x, y, z)if(le(s(x), y, s(z)), s(x), y, s(z))
if(second, x, y, z)if(le(s(x), s(y), z), s(x), s(y), z)

Original Signature

Termination of terms over the following signature is verified: greater, 0, le, s, second, if, triple, false, true, double, first

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

greater#(s(x), s(y))greater#(x, y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

le(0, y, z)greater(y, z)le(s(x), 0, z)false
le(s(x), s(y), 0)falsele(s(x), s(y), s(z))le(x, y, z)
greater(x, 0)firstgreater(0, s(y))second
greater(s(x), s(y))greater(x, y)double(0)0
double(s(x))s(s(double(x)))triple(x)if(le(x, x, double(x)), x, 0, 0)
if(false, x, y, z)trueif(first, x, y, z)if(le(s(x), y, s(z)), s(x), y, s(z))
if(second, x, y, z)if(le(s(x), s(y), z), s(x), s(y), z)

Original Signature

Termination of terms over the following signature is verified: greater, 0, le, s, second, if, triple, false, true, double, first

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

le(0, y, z)greater(y, z)le(s(x), 0, z)false
le(s(x), s(y), 0)falsele(s(x), s(y), s(z))le(x, y, z)
greater(x, 0)firstgreater(0, s(y))second
greater(s(x), s(y))greater(x, y)double(0)0
double(s(x))s(s(double(x)))triple(x)if(le(x, x, double(x)), x, 0, 0)
if(false, x, y, z)trueif(first, x, y, z)if(le(s(x), y, s(z)), s(x), y, s(z))
if(second, x, y, z)if(le(s(x), s(y), z), s(x), s(y), z)

Original Signature

Termination of terms over the following signature is verified: greater, 0, le, s, second, if, triple, false, true, double, first

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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