TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60000 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (874ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (2ms), DependencyGraph (6ms), PolynomialLinearRange4iUR (2976ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (timeout), DependencyGraph (2ms), ReductionPairSAT (3019ms), DependencyGraph (44ms), SizeChangePrinciple (timeout)].
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 | – Problem 5 was processed with processor SubtermCriterion (3ms).
 | – Problem 6 was processed with processor SubtermCriterion (1ms).
 | – Problem 7 was processed with processor SubtermCriterion (0ms).
 | – Problem 8 was processed with processor SubtermCriterion (1ms).
 | – Problem 9 was processed with processor SubtermCriterion (1ms).
 | – Problem 10 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

top#(mark(X))top#(proper(X))top#(ok(X))top#(active(X))

Rewrite Rules

active(nats)mark(adx(zeros))active(zeros)mark(cons(0, zeros))
active(incr(cons(X, Y)))mark(cons(s(X), incr(Y)))active(adx(cons(X, Y)))mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y)))mark(X)active(tl(cons(X, Y)))mark(Y)
active(adx(X))adx(active(X))active(incr(X))incr(active(X))
active(hd(X))hd(active(X))active(tl(X))tl(active(X))
adx(mark(X))mark(adx(X))incr(mark(X))mark(incr(X))
hd(mark(X))mark(hd(X))tl(mark(X))mark(tl(X))
proper(nats)ok(nats)proper(adx(X))adx(proper(X))
proper(zeros)ok(zeros)proper(cons(X1, X2))cons(proper(X1), proper(X2))
proper(0)ok(0)proper(incr(X))incr(proper(X))
proper(s(X))s(proper(X))proper(hd(X))hd(proper(X))
proper(tl(X))tl(proper(X))adx(ok(X))ok(adx(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))incr(ok(X))ok(incr(X))
s(ok(X))ok(s(X))hd(ok(X))ok(hd(X))
tl(ok(X))ok(tl(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: tl, mark, hd, nats, 0, s, zeros, adx, active, ok, incr, proper, top, cons


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

proper#(cons(X1, X2))proper#(X1)active#(adx(cons(X, Y)))cons#(X, adx(Y))
active#(adx(X))active#(X)top#(ok(X))top#(active(X))
active#(incr(cons(X, Y)))cons#(s(X), incr(Y))incr#(ok(X))incr#(X)
cons#(ok(X1), ok(X2))cons#(X1, X2)tl#(ok(X))tl#(X)
proper#(incr(X))incr#(proper(X))top#(ok(X))active#(X)
active#(tl(X))tl#(active(X))proper#(adx(X))proper#(X)
proper#(incr(X))proper#(X)adx#(ok(X))adx#(X)
incr#(mark(X))incr#(X)proper#(hd(X))proper#(X)
active#(tl(X))active#(X)top#(mark(X))proper#(X)
active#(adx(cons(X, Y)))adx#(Y)active#(incr(cons(X, Y)))incr#(Y)
proper#(hd(X))hd#(proper(X))top#(mark(X))top#(proper(X))
active#(incr(X))active#(X)active#(adx(X))adx#(active(X))
active#(adx(cons(X, Y)))incr#(cons(X, adx(Y)))proper#(cons(X1, X2))proper#(X2)
hd#(mark(X))hd#(X)active#(incr(X))incr#(active(X))
active#(hd(X))hd#(active(X))s#(ok(X))s#(X)
active#(nats)adx#(zeros)proper#(adx(X))adx#(proper(X))
adx#(mark(X))adx#(X)proper#(s(X))proper#(X)
proper#(cons(X1, X2))cons#(proper(X1), proper(X2))active#(incr(cons(X, Y)))s#(X)
tl#(mark(X))tl#(X)proper#(s(X))s#(proper(X))
active#(hd(X))active#(X)active#(zeros)cons#(0, zeros)
proper#(tl(X))tl#(proper(X))proper#(tl(X))proper#(X)
hd#(ok(X))hd#(X)

Rewrite Rules

active(nats)mark(adx(zeros))active(zeros)mark(cons(0, zeros))
active(incr(cons(X, Y)))mark(cons(s(X), incr(Y)))active(adx(cons(X, Y)))mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y)))mark(X)active(tl(cons(X, Y)))mark(Y)
active(adx(X))adx(active(X))active(incr(X))incr(active(X))
active(hd(X))hd(active(X))active(tl(X))tl(active(X))
adx(mark(X))mark(adx(X))incr(mark(X))mark(incr(X))
hd(mark(X))mark(hd(X))tl(mark(X))mark(tl(X))
proper(nats)ok(nats)proper(adx(X))adx(proper(X))
proper(zeros)ok(zeros)proper(cons(X1, X2))cons(proper(X1), proper(X2))
proper(0)ok(0)proper(incr(X))incr(proper(X))
proper(s(X))s(proper(X))proper(hd(X))hd(proper(X))
proper(tl(X))tl(proper(X))adx(ok(X))ok(adx(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))incr(ok(X))ok(incr(X))
s(ok(X))ok(s(X))hd(ok(X))ok(hd(X))
tl(ok(X))ok(tl(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: tl, mark, hd, nats, 0, s, zeros, adx, active, ok, incr, proper, cons, top

Strategy


The following SCCs where found

tl#(mark(X)) → tl#(X)tl#(ok(X)) → tl#(X)

proper#(s(X)) → proper#(X)proper#(adx(X)) → proper#(X)
proper#(cons(X1, X2)) → proper#(X1)proper#(cons(X1, X2)) → proper#(X2)
proper#(incr(X)) → proper#(X)proper#(hd(X)) → proper#(X)
proper#(tl(X)) → proper#(X)

hd#(mark(X)) → hd#(X)hd#(ok(X)) → hd#(X)

active#(incr(X)) → active#(X)active#(adx(X)) → active#(X)
active#(hd(X)) → active#(X)active#(tl(X)) → active#(X)

incr#(ok(X)) → incr#(X)incr#(mark(X)) → incr#(X)

cons#(ok(X1), ok(X2)) → cons#(X1, X2)

adx#(mark(X)) → adx#(X)adx#(ok(X)) → adx#(X)

s#(ok(X)) → s#(X)

top#(mark(X)) → top#(proper(X))top#(ok(X)) → top#(active(X))

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

incr#(ok(X))incr#(X)incr#(mark(X))incr#(X)

Rewrite Rules

active(nats)mark(adx(zeros))active(zeros)mark(cons(0, zeros))
active(incr(cons(X, Y)))mark(cons(s(X), incr(Y)))active(adx(cons(X, Y)))mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y)))mark(X)active(tl(cons(X, Y)))mark(Y)
active(adx(X))adx(active(X))active(incr(X))incr(active(X))
active(hd(X))hd(active(X))active(tl(X))tl(active(X))
adx(mark(X))mark(adx(X))incr(mark(X))mark(incr(X))
hd(mark(X))mark(hd(X))tl(mark(X))mark(tl(X))
proper(nats)ok(nats)proper(adx(X))adx(proper(X))
proper(zeros)ok(zeros)proper(cons(X1, X2))cons(proper(X1), proper(X2))
proper(0)ok(0)proper(incr(X))incr(proper(X))
proper(s(X))s(proper(X))proper(hd(X))hd(proper(X))
proper(tl(X))tl(proper(X))adx(ok(X))ok(adx(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))incr(ok(X))ok(incr(X))
s(ok(X))ok(s(X))hd(ok(X))ok(hd(X))
tl(ok(X))ok(tl(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: tl, mark, hd, nats, 0, s, zeros, adx, active, ok, incr, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

incr#(ok(X))incr#(X)incr#(mark(X))incr#(X)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

active#(incr(X))active#(X)active#(adx(X))active#(X)
active#(hd(X))active#(X)active#(tl(X))active#(X)

Rewrite Rules

active(nats)mark(adx(zeros))active(zeros)mark(cons(0, zeros))
active(incr(cons(X, Y)))mark(cons(s(X), incr(Y)))active(adx(cons(X, Y)))mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y)))mark(X)active(tl(cons(X, Y)))mark(Y)
active(adx(X))adx(active(X))active(incr(X))incr(active(X))
active(hd(X))hd(active(X))active(tl(X))tl(active(X))
adx(mark(X))mark(adx(X))incr(mark(X))mark(incr(X))
hd(mark(X))mark(hd(X))tl(mark(X))mark(tl(X))
proper(nats)ok(nats)proper(adx(X))adx(proper(X))
proper(zeros)ok(zeros)proper(cons(X1, X2))cons(proper(X1), proper(X2))
proper(0)ok(0)proper(incr(X))incr(proper(X))
proper(s(X))s(proper(X))proper(hd(X))hd(proper(X))
proper(tl(X))tl(proper(X))adx(ok(X))ok(adx(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))incr(ok(X))ok(incr(X))
s(ok(X))ok(s(X))hd(ok(X))ok(hd(X))
tl(ok(X))ok(tl(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: tl, mark, hd, nats, 0, s, zeros, adx, active, ok, incr, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

active#(incr(X))active#(X)active#(adx(X))active#(X)
active#(hd(X))active#(X)active#(tl(X))active#(X)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

adx#(mark(X))adx#(X)adx#(ok(X))adx#(X)

Rewrite Rules

active(nats)mark(adx(zeros))active(zeros)mark(cons(0, zeros))
active(incr(cons(X, Y)))mark(cons(s(X), incr(Y)))active(adx(cons(X, Y)))mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y)))mark(X)active(tl(cons(X, Y)))mark(Y)
active(adx(X))adx(active(X))active(incr(X))incr(active(X))
active(hd(X))hd(active(X))active(tl(X))tl(active(X))
adx(mark(X))mark(adx(X))incr(mark(X))mark(incr(X))
hd(mark(X))mark(hd(X))tl(mark(X))mark(tl(X))
proper(nats)ok(nats)proper(adx(X))adx(proper(X))
proper(zeros)ok(zeros)proper(cons(X1, X2))cons(proper(X1), proper(X2))
proper(0)ok(0)proper(incr(X))incr(proper(X))
proper(s(X))s(proper(X))proper(hd(X))hd(proper(X))
proper(tl(X))tl(proper(X))adx(ok(X))ok(adx(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))incr(ok(X))ok(incr(X))
s(ok(X))ok(s(X))hd(ok(X))ok(hd(X))
tl(ok(X))ok(tl(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: tl, mark, hd, nats, 0, s, zeros, adx, active, ok, incr, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

adx#(mark(X))adx#(X)adx#(ok(X))adx#(X)

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

hd#(mark(X))hd#(X)hd#(ok(X))hd#(X)

Rewrite Rules

active(nats)mark(adx(zeros))active(zeros)mark(cons(0, zeros))
active(incr(cons(X, Y)))mark(cons(s(X), incr(Y)))active(adx(cons(X, Y)))mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y)))mark(X)active(tl(cons(X, Y)))mark(Y)
active(adx(X))adx(active(X))active(incr(X))incr(active(X))
active(hd(X))hd(active(X))active(tl(X))tl(active(X))
adx(mark(X))mark(adx(X))incr(mark(X))mark(incr(X))
hd(mark(X))mark(hd(X))tl(mark(X))mark(tl(X))
proper(nats)ok(nats)proper(adx(X))adx(proper(X))
proper(zeros)ok(zeros)proper(cons(X1, X2))cons(proper(X1), proper(X2))
proper(0)ok(0)proper(incr(X))incr(proper(X))
proper(s(X))s(proper(X))proper(hd(X))hd(proper(X))
proper(tl(X))tl(proper(X))adx(ok(X))ok(adx(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))incr(ok(X))ok(incr(X))
s(ok(X))ok(s(X))hd(ok(X))ok(hd(X))
tl(ok(X))ok(tl(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: tl, mark, hd, nats, 0, s, zeros, adx, active, ok, incr, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

hd#(mark(X))hd#(X)hd#(ok(X))hd#(X)

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

tl#(mark(X))tl#(X)tl#(ok(X))tl#(X)

Rewrite Rules

active(nats)mark(adx(zeros))active(zeros)mark(cons(0, zeros))
active(incr(cons(X, Y)))mark(cons(s(X), incr(Y)))active(adx(cons(X, Y)))mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y)))mark(X)active(tl(cons(X, Y)))mark(Y)
active(adx(X))adx(active(X))active(incr(X))incr(active(X))
active(hd(X))hd(active(X))active(tl(X))tl(active(X))
adx(mark(X))mark(adx(X))incr(mark(X))mark(incr(X))
hd(mark(X))mark(hd(X))tl(mark(X))mark(tl(X))
proper(nats)ok(nats)proper(adx(X))adx(proper(X))
proper(zeros)ok(zeros)proper(cons(X1, X2))cons(proper(X1), proper(X2))
proper(0)ok(0)proper(incr(X))incr(proper(X))
proper(s(X))s(proper(X))proper(hd(X))hd(proper(X))
proper(tl(X))tl(proper(X))adx(ok(X))ok(adx(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))incr(ok(X))ok(incr(X))
s(ok(X))ok(s(X))hd(ok(X))ok(hd(X))
tl(ok(X))ok(tl(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: tl, mark, hd, nats, 0, s, zeros, adx, active, ok, incr, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

tl#(mark(X))tl#(X)tl#(ok(X))tl#(X)

Problem 8: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

s#(ok(X))s#(X)

Rewrite Rules

active(nats)mark(adx(zeros))active(zeros)mark(cons(0, zeros))
active(incr(cons(X, Y)))mark(cons(s(X), incr(Y)))active(adx(cons(X, Y)))mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y)))mark(X)active(tl(cons(X, Y)))mark(Y)
active(adx(X))adx(active(X))active(incr(X))incr(active(X))
active(hd(X))hd(active(X))active(tl(X))tl(active(X))
adx(mark(X))mark(adx(X))incr(mark(X))mark(incr(X))
hd(mark(X))mark(hd(X))tl(mark(X))mark(tl(X))
proper(nats)ok(nats)proper(adx(X))adx(proper(X))
proper(zeros)ok(zeros)proper(cons(X1, X2))cons(proper(X1), proper(X2))
proper(0)ok(0)proper(incr(X))incr(proper(X))
proper(s(X))s(proper(X))proper(hd(X))hd(proper(X))
proper(tl(X))tl(proper(X))adx(ok(X))ok(adx(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))incr(ok(X))ok(incr(X))
s(ok(X))ok(s(X))hd(ok(X))ok(hd(X))
tl(ok(X))ok(tl(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: tl, mark, hd, nats, 0, s, zeros, adx, active, ok, incr, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

s#(ok(X))s#(X)

Problem 9: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

cons#(ok(X1), ok(X2))cons#(X1, X2)

Rewrite Rules

active(nats)mark(adx(zeros))active(zeros)mark(cons(0, zeros))
active(incr(cons(X, Y)))mark(cons(s(X), incr(Y)))active(adx(cons(X, Y)))mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y)))mark(X)active(tl(cons(X, Y)))mark(Y)
active(adx(X))adx(active(X))active(incr(X))incr(active(X))
active(hd(X))hd(active(X))active(tl(X))tl(active(X))
adx(mark(X))mark(adx(X))incr(mark(X))mark(incr(X))
hd(mark(X))mark(hd(X))tl(mark(X))mark(tl(X))
proper(nats)ok(nats)proper(adx(X))adx(proper(X))
proper(zeros)ok(zeros)proper(cons(X1, X2))cons(proper(X1), proper(X2))
proper(0)ok(0)proper(incr(X))incr(proper(X))
proper(s(X))s(proper(X))proper(hd(X))hd(proper(X))
proper(tl(X))tl(proper(X))adx(ok(X))ok(adx(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))incr(ok(X))ok(incr(X))
s(ok(X))ok(s(X))hd(ok(X))ok(hd(X))
tl(ok(X))ok(tl(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: tl, mark, hd, nats, 0, s, zeros, adx, active, ok, incr, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

cons#(ok(X1), ok(X2))cons#(X1, X2)

Problem 10: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

proper#(s(X))proper#(X)proper#(adx(X))proper#(X)
proper#(cons(X1, X2))proper#(X1)proper#(cons(X1, X2))proper#(X2)
proper#(incr(X))proper#(X)proper#(hd(X))proper#(X)
proper#(tl(X))proper#(X)

Rewrite Rules

active(nats)mark(adx(zeros))active(zeros)mark(cons(0, zeros))
active(incr(cons(X, Y)))mark(cons(s(X), incr(Y)))active(adx(cons(X, Y)))mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y)))mark(X)active(tl(cons(X, Y)))mark(Y)
active(adx(X))adx(active(X))active(incr(X))incr(active(X))
active(hd(X))hd(active(X))active(tl(X))tl(active(X))
adx(mark(X))mark(adx(X))incr(mark(X))mark(incr(X))
hd(mark(X))mark(hd(X))tl(mark(X))mark(tl(X))
proper(nats)ok(nats)proper(adx(X))adx(proper(X))
proper(zeros)ok(zeros)proper(cons(X1, X2))cons(proper(X1), proper(X2))
proper(0)ok(0)proper(incr(X))incr(proper(X))
proper(s(X))s(proper(X))proper(hd(X))hd(proper(X))
proper(tl(X))tl(proper(X))adx(ok(X))ok(adx(X))
cons(ok(X1), ok(X2))ok(cons(X1, X2))incr(ok(X))ok(incr(X))
s(ok(X))ok(s(X))hd(ok(X))ok(hd(X))
tl(ok(X))ok(tl(X))top(mark(X))top(proper(X))
top(ok(X))top(active(X))

Original Signature

Termination of terms over the following signature is verified: tl, mark, hd, nats, 0, s, zeros, adx, active, ok, incr, proper, cons, top

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

proper#(adx(X))proper#(X)proper#(s(X))proper#(X)
proper#(cons(X1, X2))proper#(X1)proper#(cons(X1, X2))proper#(X2)
proper#(incr(X))proper#(X)proper#(hd(X))proper#(X)
proper#(tl(X))proper#(X)