TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60127 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (165ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – 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 (4ms).
 | – Problem 7 was processed with processor SubtermCriterion (1ms).
 | – Problem 8 was processed with processor SubtermCriterion (0ms).
 | – Problem 9 was processed with processor PolynomialLinearRange4iUR (436ms).
 |    | – Problem 11 was processed with processor BackwardInstantiation (3ms).
 |    |    | – Problem 12 remains open; application of the following processors failed [ForwardInstantiation (4ms), Propagation (1ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (2ms), Propagation (0ms)].
 | – Problem 10 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 11

Dependency Pairs

ify#(true, x, y)if#(ge(x, y), x, y)div#(x, y)ify#(ge(y, s(0)), x, y)
if#(true, x, y)div#(minus(x, y), y)

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, ify, if, div, divByZeroError, true, false, ge


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

ify#(true, x, y)ge#(x, y)div#(plus(x, y), z)div#(x, z)
ge#(s(x), 0)ge#(x, 0)minus#(0, s(x))minus#(0, x)
minus#(s(x), 0)minus#(x, 0)ify#(true, x, y)if#(ge(x, y), x, y)
div#(plus(x, y), z)div#(y, z)ge#(0, s(s(x)))ge#(0, s(x))
div#(plus(x, y), z)plus#(div(x, z), div(y, z))div#(x, y)ify#(ge(y, s(0)), x, y)
plus#(s(x), y)plus#(x, y)div#(x, y)ge#(y, s(0))
minus#(s(x), s(y))minus#(x, y)ge#(s(x), s(y))ge#(x, y)
if#(true, x, y)minus#(x, y)if#(true, x, y)div#(minus(x, y), y)
plus#(0, s(x))plus#(0, x)

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, ify, if, div, divByZeroError, true, false, ge

Strategy


The following SCCs where found

ge#(s(x), 0) → ge#(x, 0)

ge#(0, s(s(x))) → ge#(0, s(x))

plus#(s(x), y) → plus#(x, y)

minus#(0, s(x)) → minus#(0, x)

minus#(s(x), s(y)) → minus#(x, y)

ge#(s(x), s(y)) → ge#(x, y)

plus#(0, s(x)) → plus#(0, x)

minus#(s(x), 0) → minus#(x, 0)

ify#(true, x, y) → if#(ge(x, y), x, y)div#(plus(x, y), z) → div#(x, z)
div#(plus(x, y), z) → div#(y, z)div#(x, y) → ify#(ge(y, s(0)), x, y)
if#(true, x, y) → div#(minus(x, y), y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, ify, if, div, divByZeroError, true, false, ge

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

minus#(s(x), 0)minus#(x, 0)

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, ify, if, div, divByZeroError, true, false, ge

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

minus#(s(x), 0)minus#(x, 0)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, ify, if, div, divByZeroError, true, false, 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), 0)ge#(x, 0)

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, ify, if, div, divByZeroError, true, false, ge

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

ge#(s(x), 0)ge#(x, 0)

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

plus#(0, s(x))plus#(0, x)

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, ify, if, div, divByZeroError, true, false, ge

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

plus#(0, s(x))plus#(0, x)

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

minus#(0, s(x))minus#(0, x)

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, ify, if, div, divByZeroError, true, false, ge

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

minus#(0, s(x))minus#(0, x)

Problem 8: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

ge#(s(x), s(y))ge#(x, y)

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, ify, if, div, divByZeroError, true, false, ge

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

ge#(s(x), s(y))ge#(x, y)

Problem 9: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

ify#(true, x, y)if#(ge(x, y), x, y)div#(plus(x, y), z)div#(x, z)
div#(plus(x, y), z)div#(y, z)div#(x, y)ify#(ge(y, s(0)), x, y)
if#(true, x, y)div#(minus(x, y), y)

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, ify, if, div, divByZeroError, true, false, ge

Strategy


Polynomial Interpretation

Improved Usable rules

minus(s(x), 0)s(minus(x, 0))minus(s(x), s(y))minus(x, y)
minus(0, s(x))minus(0, x)minus(0, 0)0
ge(0, s(0))falsege(0, 0)true
ge(0, s(s(x)))ge(0, s(x))ge(s(x), 0)ge(x, 0)
ge(s(x), s(y))ge(x, y)

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

div#(plus(x, y), z)div#(x, z)div#(plus(x, y), z)div#(y, z)

Problem 11: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

ify#(true, x, y)if#(ge(x, y), x, y)div#(x, y)ify#(ge(y, s(0)), x, y)
if#(true, x, y)div#(minus(x, y), y)

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, minus, 0, s, if, ify, divByZeroError, div, false, true, ge

Strategy


Instantiation

For all potential predecessors l → r of the rule div#(x, y) → ify#(ge(y, s(0)), x, y) on dependency pair chains it holds that: Thus, div#(x, y) → ify#(ge(y, s(0)), x, y) is replaced by instances determined through the above matching. These instances are:
div#(minus(_x, _y), _y) → ify#(ge(_y, s(0)), minus(_x, _y), _y)

Problem 10: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

ge#(0, s(s(x)))ge#(0, s(x))

Rewrite Rules

ge(0, 0)truege(s(x), 0)ge(x, 0)
ge(0, s(0))falsege(0, s(s(x)))ge(0, s(x))
ge(s(x), s(y))ge(x, y)minus(0, 0)0
minus(0, s(x))minus(0, x)minus(s(x), 0)s(minus(x, 0))
minus(s(x), s(y))minus(x, y)plus(0, 0)0
plus(0, s(x))s(plus(0, x))plus(s(x), y)s(plus(x, y))
div(x, y)ify(ge(y, s(0)), x, y)ify(false, x, y)divByZeroError
ify(true, x, y)if(ge(x, y), x, y)if(false, x, y)0
if(true, x, y)s(div(minus(x, y), y))div(plus(x, y), z)plus(div(x, z), div(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, ify, if, div, divByZeroError, true, false, ge

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

ge#(0, s(s(x)))ge#(0, s(x))