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)