TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60021 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (170ms).
 | – Problem 2 was processed with processor SubtermCriterion (2ms).
 | – Problem 3 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (6ms), PolynomialLinearRange4iUR (83ms), DependencyGraph (5ms), PolynomialLinearRange8NegiUR (68ms), DependencyGraph (4ms), ReductionPairSAT (385ms)].
 | – Problem 4 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (3ms), PolynomialLinearRange4iUR (209ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (809ms), DependencyGraph (2ms), ReductionPairSAT (331ms)].
 | – Problem 5 was processed with processor SubtermCriterion (1ms).
 | – Problem 6 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (4ms), PolynomialLinearRange4iUR (548ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (2162ms), DependencyGraph (2ms), ReductionPairSAT (timeout)].
 | – Problem 7 was processed with processor SubtermCriterion (0ms).
 | – Problem 8 was processed with processor SubtermCriterion (5ms).

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

f2#(x, 1, z)f0#(x, z, z)f0#(0, y, x)f1#(x, y, x)
f1#(x, y, z)f2#(x, y, z)

Rewrite Rules

plus(x, y)ifPlus(isZero(x), x, inc(y))ifPlus(true, x, y)p(y)
ifPlus(false, x, y)plus(p(x), y)times(x, y)timesIter(0, x, y, 0)
timesIter(i, x, y, z)ifTimes(ge(i, x), i, x, y, z)ifTimes(true, i, x, y, z)z
ifTimes(false, i, x, y, z)timesIter(inc(i), x, y, plus(z, y))isZero(0)true
isZero(s(0))falseisZero(s(s(x)))isZero(s(x))
inc(0)s(0)inc(s(x))s(inc(x))
inc(x)s(x)p(0)0
p(s(x))xp(s(s(x)))s(p(s(x)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)f0(0, y, x)f1(x, y, x)
f1(x, y, z)f2(x, y, z)f2(x, 1, z)f0(x, z, z)
f0(x, y, z)df1(x, y, z)c

Original Signature

Termination of terms over the following signature is verified: plus, ifTimes, d, c, true, ge, ifPlus, 1, timesIter, 0, inc, s, times, p, false, f1, f0, f2, isZero




Open Dependency Pair Problem 4

Dependency Pairs

ifPlus#(false, x, y)plus#(p(x), y)plus#(x, y)ifPlus#(isZero(x), x, inc(y))

Rewrite Rules

plus(x, y)ifPlus(isZero(x), x, inc(y))ifPlus(true, x, y)p(y)
ifPlus(false, x, y)plus(p(x), y)times(x, y)timesIter(0, x, y, 0)
timesIter(i, x, y, z)ifTimes(ge(i, x), i, x, y, z)ifTimes(true, i, x, y, z)z
ifTimes(false, i, x, y, z)timesIter(inc(i), x, y, plus(z, y))isZero(0)true
isZero(s(0))falseisZero(s(s(x)))isZero(s(x))
inc(0)s(0)inc(s(x))s(inc(x))
inc(x)s(x)p(0)0
p(s(x))xp(s(s(x)))s(p(s(x)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)f0(0, y, x)f1(x, y, x)
f1(x, y, z)f2(x, y, z)f2(x, 1, z)f0(x, z, z)
f0(x, y, z)df1(x, y, z)c

Original Signature

Termination of terms over the following signature is verified: plus, ifTimes, d, c, true, ge, ifPlus, 1, timesIter, 0, inc, s, times, p, false, f1, f0, f2, isZero




Open Dependency Pair Problem 6

Dependency Pairs

ifTimes#(false, i, x, y, z)timesIter#(inc(i), x, y, plus(z, y))timesIter#(i, x, y, z)ifTimes#(ge(i, x), i, x, y, z)

Rewrite Rules

plus(x, y)ifPlus(isZero(x), x, inc(y))ifPlus(true, x, y)p(y)
ifPlus(false, x, y)plus(p(x), y)times(x, y)timesIter(0, x, y, 0)
timesIter(i, x, y, z)ifTimes(ge(i, x), i, x, y, z)ifTimes(true, i, x, y, z)z
ifTimes(false, i, x, y, z)timesIter(inc(i), x, y, plus(z, y))isZero(0)true
isZero(s(0))falseisZero(s(s(x)))isZero(s(x))
inc(0)s(0)inc(s(x))s(inc(x))
inc(x)s(x)p(0)0
p(s(x))xp(s(s(x)))s(p(s(x)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)f0(0, y, x)f1(x, y, x)
f1(x, y, z)f2(x, y, z)f2(x, 1, z)f0(x, z, z)
f0(x, y, z)df1(x, y, z)c

Original Signature

Termination of terms over the following signature is verified: plus, ifTimes, d, c, true, ge, ifPlus, 1, timesIter, 0, inc, s, times, p, false, f1, f0, f2, isZero


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

ifPlus#(false, x, y)p#(x)timesIter#(i, x, y, z)ge#(i, x)
f2#(x, 1, z)f0#(x, z, z)ifTimes#(false, i, x, y, z)timesIter#(inc(i), x, y, plus(z, y))
ifTimes#(false, i, x, y, z)plus#(z, y)plus#(x, y)inc#(y)
ifPlus#(true, x, y)p#(y)f0#(0, y, x)f1#(x, y, x)
plus#(x, y)ifPlus#(isZero(x), x, inc(y))f1#(x, y, z)f2#(x, y, z)
ifPlus#(false, x, y)plus#(p(x), y)isZero#(s(s(x)))isZero#(s(x))
ifTimes#(false, i, x, y, z)inc#(i)ge#(s(x), s(y))ge#(x, y)
timesIter#(i, x, y, z)ifTimes#(ge(i, x), i, x, y, z)times#(x, y)timesIter#(0, x, y, 0)
inc#(s(x))inc#(x)plus#(x, y)isZero#(x)
p#(s(s(x)))p#(s(x))

Rewrite Rules

plus(x, y)ifPlus(isZero(x), x, inc(y))ifPlus(true, x, y)p(y)
ifPlus(false, x, y)plus(p(x), y)times(x, y)timesIter(0, x, y, 0)
timesIter(i, x, y, z)ifTimes(ge(i, x), i, x, y, z)ifTimes(true, i, x, y, z)z
ifTimes(false, i, x, y, z)timesIter(inc(i), x, y, plus(z, y))isZero(0)true
isZero(s(0))falseisZero(s(s(x)))isZero(s(x))
inc(0)s(0)inc(s(x))s(inc(x))
inc(x)s(x)p(0)0
p(s(x))xp(s(s(x)))s(p(s(x)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)f0(0, y, x)f1(x, y, x)
f1(x, y, z)f2(x, y, z)f2(x, 1, z)f0(x, z, z)
f0(x, y, z)df1(x, y, z)c

Original Signature

Termination of terms over the following signature is verified: plus, ifTimes, d, c, true, ge, ifPlus, 1, timesIter, 0, inc, s, times, p, false, f1, f0, f2, isZero

Strategy


The following SCCs where found

ifTimes#(false, i, x, y, z) → timesIter#(inc(i), x, y, plus(z, y))timesIter#(i, x, y, z) → ifTimes#(ge(i, x), i, x, y, z)

isZero#(s(s(x))) → isZero#(s(x))

ifPlus#(false, x, y) → plus#(p(x), y)plus#(x, y) → ifPlus#(isZero(x), x, inc(y))

ge#(s(x), s(y)) → ge#(x, y)

inc#(s(x)) → inc#(x)

p#(s(s(x))) → p#(s(x))

f2#(x, 1, z) → f0#(x, z, z)f0#(0, y, x) → f1#(x, y, x)
f1#(x, y, z) → f2#(x, y, z)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

isZero#(s(s(x)))isZero#(s(x))

Rewrite Rules

plus(x, y)ifPlus(isZero(x), x, inc(y))ifPlus(true, x, y)p(y)
ifPlus(false, x, y)plus(p(x), y)times(x, y)timesIter(0, x, y, 0)
timesIter(i, x, y, z)ifTimes(ge(i, x), i, x, y, z)ifTimes(true, i, x, y, z)z
ifTimes(false, i, x, y, z)timesIter(inc(i), x, y, plus(z, y))isZero(0)true
isZero(s(0))falseisZero(s(s(x)))isZero(s(x))
inc(0)s(0)inc(s(x))s(inc(x))
inc(x)s(x)p(0)0
p(s(x))xp(s(s(x)))s(p(s(x)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)f0(0, y, x)f1(x, y, x)
f1(x, y, z)f2(x, y, z)f2(x, 1, z)f0(x, z, z)
f0(x, y, z)df1(x, y, z)c

Original Signature

Termination of terms over the following signature is verified: plus, ifTimes, d, c, true, ge, ifPlus, 1, timesIter, 0, inc, s, times, p, false, f1, f0, f2, isZero

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

isZero#(s(s(x)))isZero#(s(x))

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

inc#(s(x))inc#(x)

Rewrite Rules

plus(x, y)ifPlus(isZero(x), x, inc(y))ifPlus(true, x, y)p(y)
ifPlus(false, x, y)plus(p(x), y)times(x, y)timesIter(0, x, y, 0)
timesIter(i, x, y, z)ifTimes(ge(i, x), i, x, y, z)ifTimes(true, i, x, y, z)z
ifTimes(false, i, x, y, z)timesIter(inc(i), x, y, plus(z, y))isZero(0)true
isZero(s(0))falseisZero(s(s(x)))isZero(s(x))
inc(0)s(0)inc(s(x))s(inc(x))
inc(x)s(x)p(0)0
p(s(x))xp(s(s(x)))s(p(s(x)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)f0(0, y, x)f1(x, y, x)
f1(x, y, z)f2(x, y, z)f2(x, 1, z)f0(x, z, z)
f0(x, y, z)df1(x, y, z)c

Original Signature

Termination of terms over the following signature is verified: plus, ifTimes, d, c, true, ge, ifPlus, 1, timesIter, 0, inc, s, times, p, false, f1, f0, f2, isZero

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

inc#(s(x))inc#(x)

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

plus(x, y)ifPlus(isZero(x), x, inc(y))ifPlus(true, x, y)p(y)
ifPlus(false, x, y)plus(p(x), y)times(x, y)timesIter(0, x, y, 0)
timesIter(i, x, y, z)ifTimes(ge(i, x), i, x, y, z)ifTimes(true, i, x, y, z)z
ifTimes(false, i, x, y, z)timesIter(inc(i), x, y, plus(z, y))isZero(0)true
isZero(s(0))falseisZero(s(s(x)))isZero(s(x))
inc(0)s(0)inc(s(x))s(inc(x))
inc(x)s(x)p(0)0
p(s(x))xp(s(s(x)))s(p(s(x)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)f0(0, y, x)f1(x, y, x)
f1(x, y, z)f2(x, y, z)f2(x, 1, z)f0(x, z, z)
f0(x, y, z)df1(x, y, z)c

Original Signature

Termination of terms over the following signature is verified: plus, ifTimes, d, c, true, ge, ifPlus, 1, timesIter, 0, inc, s, times, p, false, f1, f0, f2, isZero

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 8: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

p#(s(s(x)))p#(s(x))

Rewrite Rules

plus(x, y)ifPlus(isZero(x), x, inc(y))ifPlus(true, x, y)p(y)
ifPlus(false, x, y)plus(p(x), y)times(x, y)timesIter(0, x, y, 0)
timesIter(i, x, y, z)ifTimes(ge(i, x), i, x, y, z)ifTimes(true, i, x, y, z)z
ifTimes(false, i, x, y, z)timesIter(inc(i), x, y, plus(z, y))isZero(0)true
isZero(s(0))falseisZero(s(s(x)))isZero(s(x))
inc(0)s(0)inc(s(x))s(inc(x))
inc(x)s(x)p(0)0
p(s(x))xp(s(s(x)))s(p(s(x)))
ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)f0(0, y, x)f1(x, y, x)
f1(x, y, z)f2(x, y, z)f2(x, 1, z)f0(x, z, z)
f0(x, y, z)df1(x, y, z)c

Original Signature

Termination of terms over the following signature is verified: plus, ifTimes, d, c, true, ge, ifPlus, 1, timesIter, 0, inc, s, times, p, false, f1, f0, f2, isZero

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

p#(s(s(x)))p#(s(x))