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)