YES

The TRS could be proven terminating. The proof took 1099 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (291ms).
 | – Problem 2 was processed with processor SubtermCriterion (2ms).
 |    | – Problem 7 was processed with processor DependencyGraph (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (2ms).
 |    | – Problem 8 was processed with processor PolynomialLinearRange4iUR (120ms).
 |    |    | – Problem 10 was processed with processor DependencyGraph (17ms).
 | – Problem 4 was processed with processor PolynomialLinearRange4iUR (188ms).
 |    | – Problem 9 was processed with processor PolynomialOrderingProcessor (134ms).
 | – Problem 5 was processed with processor SubtermCriterion (2ms).
 | – Problem 6 was processed with processor SubtermCriterion (2ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

divides#(y, x)div#(x, y)pr#(x, s(s(y)))divides#(s(s(y)), x)
times#(s(x), y)times#(x, y)div#(div(x, y), z)times#(y, z)
pr#(x, s(s(y)))if#(divides(s(s(y)), x), x, s(y))divides#(y, x)eq#(x, times(div(x, y), y))
quot#(s(x), s(y), z)quot#(x, y, z)times#(s(x), y)plus#(y, times(x, y))
if#(false, x, y)pr#(x, y)divides#(y, x)times#(div(x, y), y)
plus#(s(x), y)plus#(x, y)plus#(s(x), y)plus#(p(s(x)), y)
quot#(x, 0, s(z))div#(x, s(z))div#(x, y)quot#(x, y, y)
eq#(s(x), s(y))eq#(x, y)div#(div(x, y), z)div#(x, times(y, z))
prime#(s(s(x)))pr#(s(s(x)), s(x))plus#(s(x), y)p#(s(x))
plus#(x, s(y))p#(s(y))plus#(x, s(y))plus#(x, p(s(y)))

Rewrite Rules

p(s(x))xplus(x, 0)x
plus(0, y)yplus(s(x), y)s(plus(x, y))
plus(s(x), y)s(plus(p(s(x)), y))plus(x, s(y))s(plus(x, p(s(y))))
times(0, y)0times(s(0), y)y
times(s(x), y)plus(y, times(x, y))div(0, y)0
div(x, y)quot(x, y, y)quot(0, s(y), z)0
quot(s(x), s(y), z)quot(x, y, z)quot(x, 0, s(z))s(div(x, s(z)))
div(div(x, y), z)div(x, times(y, z))eq(0, 0)true
eq(s(x), 0)falseeq(0, s(y))false
eq(s(x), s(y))eq(x, y)divides(y, x)eq(x, times(div(x, y), y))
prime(s(s(x)))pr(s(s(x)), s(x))pr(x, s(0))true
pr(x, s(s(y)))if(divides(s(s(y)), x), x, s(y))if(true, x, y)false
if(false, x, y)pr(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, div, true, divides, prime, 0, s, times, if, p, false, quot, pr, eq

Strategy


The following SCCs where found

if#(false, x, y) → pr#(x, y)pr#(x, s(s(y))) → if#(divides(s(s(y)), x), x, s(y))

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

times#(s(x), y) → times#(x, y)

eq#(s(x), s(y)) → eq#(x, y)

quot#(x, 0, s(z)) → div#(x, s(z))div#(x, y) → quot#(x, y, y)
div#(div(x, y), z) → div#(x, times(y, z))quot#(s(x), s(y), z) → quot#(x, y, z)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

if#(false, x, y)pr#(x, y)pr#(x, s(s(y)))if#(divides(s(s(y)), x), x, s(y))

Rewrite Rules

p(s(x))xplus(x, 0)x
plus(0, y)yplus(s(x), y)s(plus(x, y))
plus(s(x), y)s(plus(p(s(x)), y))plus(x, s(y))s(plus(x, p(s(y))))
times(0, y)0times(s(0), y)y
times(s(x), y)plus(y, times(x, y))div(0, y)0
div(x, y)quot(x, y, y)quot(0, s(y), z)0
quot(s(x), s(y), z)quot(x, y, z)quot(x, 0, s(z))s(div(x, s(z)))
div(div(x, y), z)div(x, times(y, z))eq(0, 0)true
eq(s(x), 0)falseeq(0, s(y))false
eq(s(x), s(y))eq(x, y)divides(y, x)eq(x, times(div(x, y), y))
prime(s(s(x)))pr(s(s(x)), s(x))pr(x, s(0))true
pr(x, s(s(y)))if(divides(s(s(y)), x), x, s(y))if(true, x, y)false
if(false, x, y)pr(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, div, true, divides, prime, 0, s, times, if, p, false, quot, pr, eq

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

pr#(x, s(s(y)))if#(divides(s(s(y)), x), x, s(y))

Problem 7: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if#(false, x, y)pr#(x, y)

Rewrite Rules

p(s(x))xplus(x, 0)x
plus(0, y)yplus(s(x), y)s(plus(x, y))
plus(s(x), y)s(plus(p(s(x)), y))plus(x, s(y))s(plus(x, p(s(y))))
times(0, y)0times(s(0), y)y
times(s(x), y)plus(y, times(x, y))div(0, y)0
div(x, y)quot(x, y, y)quot(0, s(y), z)0
quot(s(x), s(y), z)quot(x, y, z)quot(x, 0, s(z))s(div(x, s(z)))
div(div(x, y), z)div(x, times(y, z))eq(0, 0)true
eq(s(x), 0)falseeq(0, s(y))false
eq(s(x), s(y))eq(x, y)divides(y, x)eq(x, times(div(x, y), y))
prime(s(s(x)))pr(s(s(x)), s(x))pr(x, s(0))true
pr(x, s(s(y)))if(divides(s(s(y)), x), x, s(y))if(true, x, y)false
if(false, x, y)pr(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, div, true, divides, prime, 0, s, times, if, p, false, quot, pr, eq

Strategy


There are no SCCs!

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

quot#(x, 0, s(z))div#(x, s(z))div#(x, y)quot#(x, y, y)
div#(div(x, y), z)div#(x, times(y, z))quot#(s(x), s(y), z)quot#(x, y, z)

Rewrite Rules

p(s(x))xplus(x, 0)x
plus(0, y)yplus(s(x), y)s(plus(x, y))
plus(s(x), y)s(plus(p(s(x)), y))plus(x, s(y))s(plus(x, p(s(y))))
times(0, y)0times(s(0), y)y
times(s(x), y)plus(y, times(x, y))div(0, y)0
div(x, y)quot(x, y, y)quot(0, s(y), z)0
quot(s(x), s(y), z)quot(x, y, z)quot(x, 0, s(z))s(div(x, s(z)))
div(div(x, y), z)div(x, times(y, z))eq(0, 0)true
eq(s(x), 0)falseeq(0, s(y))false
eq(s(x), s(y))eq(x, y)divides(y, x)eq(x, times(div(x, y), y))
prime(s(s(x)))pr(s(s(x)), s(x))pr(x, s(0))true
pr(x, s(s(y)))if(divides(s(s(y)), x), x, s(y))if(true, x, y)false
if(false, x, y)pr(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, div, true, divides, prime, 0, s, times, if, p, false, quot, pr, eq

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

div#(div(x, y), z)div#(x, times(y, z))quot#(s(x), s(y), z)quot#(x, y, z)

Problem 8: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

quot#(x, 0, s(z))div#(x, s(z))div#(x, y)quot#(x, y, y)

Rewrite Rules

p(s(x))xplus(x, 0)x
plus(0, y)yplus(s(x), y)s(plus(x, y))
plus(s(x), y)s(plus(p(s(x)), y))plus(x, s(y))s(plus(x, p(s(y))))
times(0, y)0times(s(0), y)y
times(s(x), y)plus(y, times(x, y))div(0, y)0
div(x, y)quot(x, y, y)quot(0, s(y), z)0
quot(s(x), s(y), z)quot(x, y, z)quot(x, 0, s(z))s(div(x, s(z)))
div(div(x, y), z)div(x, times(y, z))eq(0, 0)true
eq(s(x), 0)falseeq(0, s(y))false
eq(s(x), s(y))eq(x, y)divides(y, x)eq(x, times(div(x, y), y))
prime(s(s(x)))pr(s(s(x)), s(x))pr(x, s(0))true
pr(x, s(s(y)))if(divides(s(s(y)), x), x, s(y))if(true, x, y)false
if(false, x, y)pr(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, div, true, divides, prime, 0, s, times, if, p, false, quot, pr, eq

Strategy


Polynomial Interpretation

There are no usable rules

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

quot#(x, 0, s(z))div#(x, s(z))

Problem 10: DependencyGraph



Dependency Pair Problem

Dependency Pairs

div#(x, y)quot#(x, y, y)

Rewrite Rules

p(s(x))xplus(x, 0)x
plus(0, y)yplus(s(x), y)s(plus(x, y))
plus(s(x), y)s(plus(p(s(x)), y))plus(x, s(y))s(plus(x, p(s(y))))
times(0, y)0times(s(0), y)y
times(s(x), y)plus(y, times(x, y))div(0, y)0
div(x, y)quot(x, y, y)quot(0, s(y), z)0
quot(s(x), s(y), z)quot(x, y, z)quot(x, 0, s(z))s(div(x, s(z)))
div(div(x, y), z)div(x, times(y, z))eq(0, 0)true
eq(s(x), 0)falseeq(0, s(y))false
eq(s(x), s(y))eq(x, y)divides(y, x)eq(x, times(div(x, y), y))
prime(s(s(x)))pr(s(s(x)), s(x))pr(x, s(0))true
pr(x, s(s(y)))if(divides(s(s(y)), x), x, s(y))if(true, x, y)false
if(false, x, y)pr(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, div, true, divides, prime, 0, s, times, if, p, false, quot, pr, eq

Strategy


There are no SCCs!

Problem 4: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

p(s(x))xplus(x, 0)x
plus(0, y)yplus(s(x), y)s(plus(x, y))
plus(s(x), y)s(plus(p(s(x)), y))plus(x, s(y))s(plus(x, p(s(y))))
times(0, y)0times(s(0), y)y
times(s(x), y)plus(y, times(x, y))div(0, y)0
div(x, y)quot(x, y, y)quot(0, s(y), z)0
quot(s(x), s(y), z)quot(x, y, z)quot(x, 0, s(z))s(div(x, s(z)))
div(div(x, y), z)div(x, times(y, z))eq(0, 0)true
eq(s(x), 0)falseeq(0, s(y))false
eq(s(x), s(y))eq(x, y)divides(y, x)eq(x, times(div(x, y), y))
prime(s(s(x)))pr(s(s(x)), s(x))pr(x, s(0))true
pr(x, s(s(y)))if(divides(s(s(y)), x), x, s(y))if(true, x, y)false
if(false, x, y)pr(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, div, true, divides, prime, 0, s, times, if, p, false, quot, pr, eq

Strategy


Polynomial Interpretation

Improved Usable rules

p(s(x))x

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

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

Problem 9: PolynomialOrderingProcessor



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

p(s(x))xplus(x, 0)x
plus(0, y)yplus(s(x), y)s(plus(x, y))
plus(s(x), y)s(plus(p(s(x)), y))plus(x, s(y))s(plus(x, p(s(y))))
times(0, y)0times(s(0), y)y
times(s(x), y)plus(y, times(x, y))div(0, y)0
div(x, y)quot(x, y, y)quot(0, s(y), z)0
quot(s(x), s(y), z)quot(x, y, z)quot(x, 0, s(z))s(div(x, s(z)))
div(div(x, y), z)div(x, times(y, z))eq(0, 0)true
eq(s(x), 0)falseeq(0, s(y))false
eq(s(x), s(y))eq(x, y)divides(y, x)eq(x, times(div(x, y), y))
prime(s(s(x)))pr(s(s(x)), s(x))pr(x, s(0))true
pr(x, s(s(y)))if(divides(s(s(y)), x), x, s(y))if(true, x, y)false
if(false, x, y)pr(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, div, true, divides, prime, 0, s, times, if, p, false, quot, pr, eq

Strategy


Polynomial Interpretation

Improved Usable rules

p(s(x))x

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

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

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

eq#(s(x), s(y))eq#(x, y)

Rewrite Rules

p(s(x))xplus(x, 0)x
plus(0, y)yplus(s(x), y)s(plus(x, y))
plus(s(x), y)s(plus(p(s(x)), y))plus(x, s(y))s(plus(x, p(s(y))))
times(0, y)0times(s(0), y)y
times(s(x), y)plus(y, times(x, y))div(0, y)0
div(x, y)quot(x, y, y)quot(0, s(y), z)0
quot(s(x), s(y), z)quot(x, y, z)quot(x, 0, s(z))s(div(x, s(z)))
div(div(x, y), z)div(x, times(y, z))eq(0, 0)true
eq(s(x), 0)falseeq(0, s(y))false
eq(s(x), s(y))eq(x, y)divides(y, x)eq(x, times(div(x, y), y))
prime(s(s(x)))pr(s(s(x)), s(x))pr(x, s(0))true
pr(x, s(s(y)))if(divides(s(s(y)), x), x, s(y))if(true, x, y)false
if(false, x, y)pr(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, div, true, divides, prime, 0, s, times, if, p, false, quot, pr, eq

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

eq#(s(x), s(y))eq#(x, y)

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

p(s(x))xplus(x, 0)x
plus(0, y)yplus(s(x), y)s(plus(x, y))
plus(s(x), y)s(plus(p(s(x)), y))plus(x, s(y))s(plus(x, p(s(y))))
times(0, y)0times(s(0), y)y
times(s(x), y)plus(y, times(x, y))div(0, y)0
div(x, y)quot(x, y, y)quot(0, s(y), z)0
quot(s(x), s(y), z)quot(x, y, z)quot(x, 0, s(z))s(div(x, s(z)))
div(div(x, y), z)div(x, times(y, z))eq(0, 0)true
eq(s(x), 0)falseeq(0, s(y))false
eq(s(x), s(y))eq(x, y)divides(y, x)eq(x, times(div(x, y), y))
prime(s(s(x)))pr(s(s(x)), s(x))pr(x, s(0))true
pr(x, s(s(y)))if(divides(s(s(y)), x), x, s(y))if(true, x, y)false
if(false, x, y)pr(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, div, true, divides, prime, 0, s, times, if, p, false, quot, pr, eq

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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