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))