TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60020 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (175ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (4ms), PolynomialLinearRange4iUR (1618ms), DependencyGraph (4ms), PolynomialLinearRange8NegiUR (3403ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (3390ms), DependencyGraph (4ms), ReductionPairSAT (1374ms), DependencyGraph (4ms), SizeChangePrinciple (459ms), ForwardNarrowing (1ms), BackwardInstantiation (2ms), ForwardInstantiation (2ms), Propagation (1ms)].
 | – Problem 3 was processed with processor PolynomialLinearRange8NegiUR (13321ms).
 | – 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

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)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
minus(minus(x, y), z)minus(x, plus(y, z))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)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(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, quot


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)))
minus#(minus(x, y), z)plus#(y, z)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))minus#(minus(x, y), z)minus#(x, plus(y, z))
plus#(s(x), x)if#(gt(x, x), id(x), id(x))quot#(s(x), s(y))minus#(x, y)
plus#(s(x), s(y))id#(x)plus#(s(x), s(y))plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
plus#(s(x), s(y))if#(not(gt(x, y)), id(x), id(y))gt#(s(x), s(y))gt#(x, y)
quot#(s(x), s(y))quot#(minus(x, y), s(y))minus#(s(x), s(y))minus#(x, y)
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)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
minus(minus(x, y), z)minus(x, plus(y, z))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)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(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, zero, quot

Strategy


The following SCCs where found

minus#(s(x), s(y)) → minus#(x, y)minus#(minus(x, y), z) → minus#(x, plus(y, z))

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)

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

Problem 3: PolynomialLinearRange8NegiUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
minus(minus(x, y), z)minus(x, plus(y, z))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)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(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, zero, quot

Strategy


Polynomial Interpretation

Improved Usable rules

minus(s(x), s(y))minus(x, y)minus(minus(x, y), z)minus(x, plus(y, z))
minus(x, 0)x

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

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

Problem 4: 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)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
minus(minus(x, y), z)minus(x, plus(y, z))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)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(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, zero, quot

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
minus(minus(x, y), z)minus(x, plus(y, z))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)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(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, zero, quot

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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