TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60063 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (444ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (10ms), PolynomialLinearRange4iUR (1121ms), DependencyGraph (9ms), PolynomialLinearRange8NegiUR (3443ms), DependencyGraph (8ms), ReductionPairSAT (1506ms)].
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (3334ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (10046ms), DependencyGraph (1ms), ReductionPairSAT (2852ms)].
 | – Problem 5 was processed with processor SubtermCriterion (0ms).
 | – Problem 6 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (5ms), PolynomialLinearRange4iUR (21ms), DependencyGraph (4ms), PolynomialLinearRange8NegiUR (0ms), DependencyGraph (4ms), ReductionPairSAT (37109ms)].
 | – Problem 7 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

plus#(id(x), s(y))plus#(x, if(gt(s(y), y), y, s(y)))plus#(s(x), s(y))plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
plus#(s(x), y)plus#(x, y)plus#(s(x), x)plus#(if(gt(x, x), id(x), id(x)), s(x))

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
minus(minus(x, y), z)minus(x, plus(y, z))app(nil, k)k
app(l, nil)lapp(cons(x, l), k)cons(x, app(l, k))
sum(cons(x, nil))cons(x, nil)sum(cons(x, cons(y, l)))sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k))))sum(app(l, sum(cons(x, cons(y, k)))))plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))plus(zero, y)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, app, minus, true, sum, zero, id, not, 0, s, if, false, gt, quot, nil, cons




Open Dependency Pair Problem 4

Dependency Pairs

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

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
minus(minus(x, y), z)minus(x, plus(y, z))app(nil, k)k
app(l, nil)lapp(cons(x, l), k)cons(x, app(l, k))
sum(cons(x, nil))cons(x, nil)sum(cons(x, cons(y, l)))sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k))))sum(app(l, sum(cons(x, cons(y, k)))))plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))plus(zero, y)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, app, minus, true, sum, zero, id, not, 0, s, if, false, gt, quot, nil, cons




Open Dependency Pair Problem 6

Dependency Pairs

sum#(cons(x, cons(y, l)))sum#(cons(plus(x, y), l))sum#(app(l, cons(x, cons(y, k))))sum#(app(l, sum(cons(x, cons(y, k)))))
sum#(app(l, cons(x, cons(y, k))))sum#(cons(x, cons(y, k)))

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
minus(minus(x, y), z)minus(x, plus(y, z))app(nil, k)k
app(l, nil)lapp(cons(x, l), k)cons(x, app(l, k))
sum(cons(x, nil))cons(x, nil)sum(cons(x, cons(y, l)))sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k))))sum(app(l, sum(cons(x, cons(y, k)))))plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))plus(zero, y)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, app, minus, true, sum, zero, id, not, 0, s, if, false, gt, quot, nil, cons


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

plus#(s(x), s(y))gt#(x, y)plus#(s(x), x)gt#(x, x)
plus#(id(x), s(y))if#(gt(s(y), y), y, s(y))minus#(minus(x, y), z)minus#(x, plus(y, z))
plus#(s(x), x)if#(gt(x, x), id(x), id(x))quot#(s(x), s(y))minus#(x, y)
plus#(s(x), s(y))if#(not(gt(x, y)), id(x), id(y))plus#(s(x), y)plus#(x, y)
gt#(s(x), s(y))gt#(x, y)quot#(s(x), s(y))quot#(minus(x, y), s(y))
not#(x)if#(x, false, true)app#(cons(x, l), k)app#(l, k)
plus#(s(x), x)id#(x)plus#(id(x), s(y))plus#(x, if(gt(s(y), y), y, s(y)))
minus#(minus(x, y), z)plus#(y, z)sum#(cons(x, cons(y, l)))sum#(cons(plus(x, y), l))
plus#(id(x), s(y))gt#(s(y), y)plus#(s(x), s(y))if#(gt(x, y), x, y)
plus#(s(x), x)plus#(if(gt(x, x), id(x), id(x)), s(x))plus#(s(x), s(y))id#(x)
sum#(app(l, cons(x, cons(y, k))))app#(l, sum(cons(x, cons(y, k))))plus#(s(x), s(y))plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
sum#(cons(x, cons(y, l)))plus#(x, y)sum#(app(l, cons(x, cons(y, k))))sum#(cons(x, cons(y, k)))
sum#(app(l, cons(x, cons(y, k))))sum#(app(l, sum(cons(x, cons(y, k)))))minus#(s(x), s(y))minus#(x, y)
plus#(s(x), s(y))id#(y)plus#(s(x), s(y))not#(gt(x, y))

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
minus(minus(x, y), z)minus(x, plus(y, z))app(nil, k)k
app(l, nil)lapp(cons(x, l), k)cons(x, app(l, k))
sum(cons(x, nil))cons(x, nil)sum(cons(x, cons(y, l)))sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k))))sum(app(l, sum(cons(x, cons(y, k)))))plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))plus(zero, y)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, app, minus, true, sum, zero, id, not, 0, s, if, false, gt, quot, cons, nil

Strategy


The following SCCs where found

sum#(cons(x, cons(y, l))) → sum#(cons(plus(x, y), l))sum#(app(l, cons(x, cons(y, k)))) → sum#(cons(x, cons(y, k)))
sum#(app(l, cons(x, cons(y, k)))) → sum#(app(l, sum(cons(x, cons(y, k)))))

minus#(s(x), s(y)) → minus#(x, y)minus#(minus(x, y), z) → minus#(x, plus(y, z))

gt#(s(x), s(y)) → gt#(x, y)

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

app#(cons(x, l), k) → app#(l, k)

plus#(id(x), s(y)) → plus#(x, if(gt(s(y), y), y, s(y)))plus#(s(x), s(y)) → plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
plus#(s(x), y) → plus#(x, y)plus#(s(x), x) → plus#(if(gt(x, x), id(x), id(x)), s(x))

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

app#(cons(x, l), k)app#(l, k)

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
minus(minus(x, y), z)minus(x, plus(y, z))app(nil, k)k
app(l, nil)lapp(cons(x, l), k)cons(x, app(l, k))
sum(cons(x, nil))cons(x, nil)sum(cons(x, cons(y, l)))sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k))))sum(app(l, sum(cons(x, cons(y, k)))))plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))plus(zero, y)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, app, minus, true, sum, zero, id, not, 0, s, if, false, gt, quot, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

app#(cons(x, l), k)app#(l, k)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
minus(minus(x, y), z)minus(x, plus(y, z))app(nil, k)k
app(l, nil)lapp(cons(x, l), k)cons(x, app(l, k))
sum(cons(x, nil))cons(x, nil)sum(cons(x, cons(y, l)))sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k))))sum(app(l, sum(cons(x, cons(y, k)))))plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))plus(zero, y)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, app, minus, true, sum, zero, id, not, 0, s, if, false, gt, quot, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

gt#(s(x), s(y))gt#(x, y)

Rewrite Rules

minus(x, 0)xminus(s(x), s(y))minus(x, y)
quot(0, s(y))0quot(s(x), s(y))s(quot(minus(x, y), s(y)))
plus(0, y)yplus(s(x), y)s(plus(x, y))
minus(minus(x, y), z)minus(x, plus(y, z))app(nil, k)k
app(l, nil)lapp(cons(x, l), k)cons(x, app(l, k))
sum(cons(x, nil))cons(x, nil)sum(cons(x, cons(y, l)))sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k))))sum(app(l, sum(cons(x, cons(y, k)))))plus(s(x), s(y))s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x)plus(if(gt(x, x), id(x), id(x)), s(x))plus(zero, y)y
plus(id(x), s(y))s(plus(x, if(gt(s(y), y), y, s(y))))id(x)x
if(true, x, y)xif(false, x, y)y
not(x)if(x, false, true)gt(s(x), zero)true
gt(zero, y)falsegt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: plus, app, minus, true, sum, zero, id, not, 0, s, if, false, gt, quot, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

gt#(s(x), s(y))gt#(x, y)