TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60003 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (61ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor BackwardInstantiation (2ms).
 |    | – Problem 6 was processed with processor BackwardInstantiation (1ms).
 |    |    | – Problem 7 was processed with processor Propagation (3ms).
 |    |    |    | – Problem 8 remains open; application of the following processors failed [ForwardNarrowing (0ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (1ms)].
 | – Problem 5 was processed with processor SubtermCriterion (0ms).

The following open problems remain:



Open Dependency Pair Problem 4

Dependency Pairs

if#(false, x, y, z)facIter#(x, z)facIter#(x, y)if#(isZero(x), minus(x, s(0)), y, times(y, x))

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
p(s(x))xp(0)0
minus(x, 0)xminus(0, x)0
minus(x, s(y))p(minus(x, y))isZero(0)true
isZero(s(x))falsefacIter(x, y)if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z)yif(false, x, y, z)facIter(x, z)
factorial(x)facIter(x, s(0))

Original Signature

Termination of terms over the following signature is verified: plus, minus, 0, s, times, if, p, false, true, factorial, facIter, isZero


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

times#(s(x), y)plus#(y, times(x, y))facIter#(x, y)minus#(x, s(0))
times#(s(x), y)times#(x, y)if#(false, x, y, z)facIter#(x, z)
plus#(s(x), y)plus#(x, y)minus#(x, s(y))minus#(x, y)
facIter#(x, y)isZero#(x)minus#(x, s(y))p#(minus(x, y))
facIter#(x, y)times#(y, x)factorial#(x)facIter#(x, s(0))
facIter#(x, y)if#(isZero(x), minus(x, s(0)), y, times(y, x))

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
p(s(x))xp(0)0
minus(x, 0)xminus(0, x)0
minus(x, s(y))p(minus(x, y))isZero(0)true
isZero(s(x))falsefacIter(x, y)if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z)yif(false, x, y, z)facIter(x, z)
factorial(x)facIter(x, s(0))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, times, if, p, true, false, facIter, factorial, isZero

Strategy


The following SCCs where found

if#(false, x, y, z) → facIter#(x, z)facIter#(x, y) → if#(isZero(x), minus(x, s(0)), y, times(y, x))

times#(s(x), y) → times#(x, y)

plus#(s(x), y) → plus#(x, y)

minus#(x, s(y)) → minus#(x, y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
p(s(x))xp(0)0
minus(x, 0)xminus(0, x)0
minus(x, s(y))p(minus(x, y))isZero(0)true
isZero(s(x))falsefacIter(x, y)if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z)yif(false, x, y, z)facIter(x, z)
factorial(x)facIter(x, s(0))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, times, if, p, true, false, facIter, factorial, isZero

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
p(s(x))xp(0)0
minus(x, 0)xminus(0, x)0
minus(x, s(y))p(minus(x, y))isZero(0)true
isZero(s(x))falsefacIter(x, y)if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z)yif(false, x, y, z)facIter(x, z)
factorial(x)facIter(x, s(0))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, times, if, p, true, false, facIter, factorial, isZero

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 4: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

if#(false, x, y, z)facIter#(x, z)facIter#(x, y)if#(isZero(x), minus(x, s(0)), y, times(y, x))

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
p(s(x))xp(0)0
minus(x, 0)xminus(0, x)0
minus(x, s(y))p(minus(x, y))isZero(0)true
isZero(s(x))falsefacIter(x, y)if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z)yif(false, x, y, z)facIter(x, z)
factorial(x)facIter(x, s(0))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, times, if, p, true, false, facIter, factorial, isZero

Strategy


Instantiation

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

Problem 6: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

facIter#(_x, _z)if#(isZero(_x), minus(_x, s(0)), _z, times(_z, _x))if#(false, x, y, z)facIter#(x, z)

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
p(s(x))xp(0)0
minus(x, 0)xminus(0, x)0
minus(x, s(y))p(minus(x, y))isZero(0)true
isZero(s(x))falsefacIter(x, y)if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z)yif(false, x, y, z)facIter(x, z)
factorial(x)facIter(x, s(0))

Original Signature

Termination of terms over the following signature is verified: plus, minus, 0, s, times, if, p, false, true, factorial, facIter, isZero

Strategy


Instantiation

For all potential predecessors l → r of the rule facIter#(_x, _z) → if#(isZero(_x), minus(_x, s(0)), _z, times(_z, _x)) on dependency pair chains it holds that: Thus, facIter#(_x, _z) → if#(isZero(_x), minus(_x, s(0)), _z, times(_z, _x)) is replaced by instances determined through the above matching. These instances are:
facIter#(x, z) → if#(isZero(x), minus(x, s(0)), z, times(z, x))

Problem 7: Propagation



Dependency Pair Problem

Dependency Pairs

if#(false, x, y, z)facIter#(x, z)facIter#(x, z)if#(isZero(x), minus(x, s(0)), z, times(z, x))

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
p(s(x))xp(0)0
minus(x, 0)xminus(0, x)0
minus(x, s(y))p(minus(x, y))isZero(0)true
isZero(s(x))falsefacIter(x, y)if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z)yif(false, x, y, z)facIter(x, z)
factorial(x)facIter(x, s(0))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, times, if, p, true, false, facIter, factorial, isZero

Strategy


The dependency pairs if#(false, x, y, z) → facIter#(x, z) and facIter#(x, z) → if#(isZero(x), minus(x, s(0)), z, times(z, x)) are consolidated into the rule if#(false, x, y, z) → if#(isZero(x), minus(x, s(0)), z, times(z, x)) .

This is possible as

The dependency pairs if#(false, x, y, z) → facIter#(x, z) and facIter#(x, z) → if#(isZero(x), minus(x, s(0)), z, times(z, x)) are consolidated into the rule if#(false, x, y, z) → if#(isZero(x), minus(x, s(0)), z, times(z, x)) .

This is possible as


Summary

Removed Dependency PairsAdded Dependency Pairs
if#(false, x, y, z) → facIter#(x, z)if#(false, x, y, z) → if#(isZero(x), minus(x, s(0)), z, times(z, x))
facIter#(x, z) → if#(isZero(x), minus(x, s(0)), z, times(z, x)) 

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

times#(s(x), y)times#(x, y)

Rewrite Rules

plus(0, x)xplus(s(x), y)s(plus(x, y))
times(0, y)0times(s(x), y)plus(y, times(x, y))
p(s(x))xp(0)0
minus(x, 0)xminus(0, x)0
minus(x, s(y))p(minus(x, y))isZero(0)true
isZero(s(x))falsefacIter(x, y)if(isZero(x), minus(x, s(0)), y, times(y, x))
if(true, x, y, z)yif(false, x, y, z)facIter(x, z)
factorial(x)facIter(x, s(0))

Original Signature

Termination of terms over the following signature is verified: plus, 0, minus, s, times, if, p, true, false, facIter, factorial, isZero

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

times#(s(x), y)times#(x, y)