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 (157ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (3ms), PolynomialLinearRange4iUR (321ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (8154ms), DependencyGraph (2ms), ReductionPairSAT (14695ms), DependencyGraph (3ms), SizeChangePrinciple (69ms)].
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (1073ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (7866ms), DependencyGraph (25ms), ReductionPairSAT (2671ms), DependencyGraph (2ms), SizeChangePrinciple (timeout)].
 | – Problem 5 was processed with processor SubtermCriterion (1ms).
 | – Problem 6 was processed with processor SubtermCriterion (0ms).
 | – Problem 7 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

if#(false, x, y, z)loop#(x, double(y), s(z))loop#(x, s(y), z)if#(le(x, s(y)), x, s(y), z)

Rewrite Rules

le(s(x), 0)falsele(0, y)true
le(s(x), s(y))le(x, y)double(0)0
double(s(x))s(s(double(x)))log(0)logError
log(s(x))loop(s(x), s(0), 0)loop(x, s(y), z)if(le(x, s(y)), x, s(y), z)
if(true, x, y, z)zif(false, x, y, z)loop(x, double(y), s(z))
maplog(xs)mapIter(xs, nil)mapIter(xs, ys)ifmap(isempty(xs), xs, ys)
ifmap(true, xs, ys)ysifmap(false, xs, ys)mapIter(droplast(xs), cons(log(last(xs)), ys))
isempty(nil)trueisempty(cons(x, xs))false
last(nil)errorlast(cons(x, nil))x
last(cons(x, cons(y, xs)))last(cons(y, xs))droplast(nil)nil
droplast(cons(x, nil))nildroplast(cons(x, cons(y, xs)))cons(x, droplast(cons(y, xs)))
abac

Original Signature

Termination of terms over the following signature is verified: b, error, last, c, a, ifmap, true, logError, double, log, 0, mapIter, le, s, isempty, if, loop, false, droplast, cons, nil, maplog




Open Dependency Pair Problem 4

Dependency Pairs

mapIter#(xs, ys)ifmap#(isempty(xs), xs, ys)ifmap#(false, xs, ys)mapIter#(droplast(xs), cons(log(last(xs)), ys))

Rewrite Rules

le(s(x), 0)falsele(0, y)true
le(s(x), s(y))le(x, y)double(0)0
double(s(x))s(s(double(x)))log(0)logError
log(s(x))loop(s(x), s(0), 0)loop(x, s(y), z)if(le(x, s(y)), x, s(y), z)
if(true, x, y, z)zif(false, x, y, z)loop(x, double(y), s(z))
maplog(xs)mapIter(xs, nil)mapIter(xs, ys)ifmap(isempty(xs), xs, ys)
ifmap(true, xs, ys)ysifmap(false, xs, ys)mapIter(droplast(xs), cons(log(last(xs)), ys))
isempty(nil)trueisempty(cons(x, xs))false
last(nil)errorlast(cons(x, nil))x
last(cons(x, cons(y, xs)))last(cons(y, xs))droplast(nil)nil
droplast(cons(x, nil))nildroplast(cons(x, cons(y, xs)))cons(x, droplast(cons(y, xs)))
abac

Original Signature

Termination of terms over the following signature is verified: b, error, last, c, a, ifmap, true, logError, double, log, 0, mapIter, le, s, isempty, if, loop, false, droplast, cons, nil, maplog


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

last#(cons(x, cons(y, xs)))last#(cons(y, xs))log#(s(x))loop#(s(x), s(0), 0)
droplast#(cons(x, cons(y, xs)))droplast#(cons(y, xs))ifmap#(false, xs, ys)mapIter#(droplast(xs), cons(log(last(xs)), ys))
loop#(x, s(y), z)if#(le(x, s(y)), x, s(y), z)ifmap#(false, xs, ys)last#(xs)
ifmap#(false, xs, ys)droplast#(xs)le#(s(x), s(y))le#(x, y)
if#(false, x, y, z)loop#(x, double(y), s(z))mapIter#(xs, ys)ifmap#(isempty(xs), xs, ys)
double#(s(x))double#(x)maplog#(xs)mapIter#(xs, nil)
ifmap#(false, xs, ys)log#(last(xs))if#(false, x, y, z)double#(y)
mapIter#(xs, ys)isempty#(xs)loop#(x, s(y), z)le#(x, s(y))

Rewrite Rules

le(s(x), 0)falsele(0, y)true
le(s(x), s(y))le(x, y)double(0)0
double(s(x))s(s(double(x)))log(0)logError
log(s(x))loop(s(x), s(0), 0)loop(x, s(y), z)if(le(x, s(y)), x, s(y), z)
if(true, x, y, z)zif(false, x, y, z)loop(x, double(y), s(z))
maplog(xs)mapIter(xs, nil)mapIter(xs, ys)ifmap(isempty(xs), xs, ys)
ifmap(true, xs, ys)ysifmap(false, xs, ys)mapIter(droplast(xs), cons(log(last(xs)), ys))
isempty(nil)trueisempty(cons(x, xs))false
last(nil)errorlast(cons(x, nil))x
last(cons(x, cons(y, xs)))last(cons(y, xs))droplast(nil)nil
droplast(cons(x, nil))nildroplast(cons(x, cons(y, xs)))cons(x, droplast(cons(y, xs)))
abac

Original Signature

Termination of terms over the following signature is verified: b, error, last, c, a, ifmap, true, logError, double, log, 0, mapIter, s, le, isempty, if, loop, false, droplast, maplog, nil, cons

Strategy


The following SCCs where found

last#(cons(x, cons(y, xs))) → last#(cons(y, xs))

le#(s(x), s(y)) → le#(x, y)

if#(false, x, y, z) → loop#(x, double(y), s(z))loop#(x, s(y), z) → if#(le(x, s(y)), x, s(y), z)

droplast#(cons(x, cons(y, xs))) → droplast#(cons(y, xs))

double#(s(x)) → double#(x)

mapIter#(xs, ys) → ifmap#(isempty(xs), xs, ys)ifmap#(false, xs, ys) → mapIter#(droplast(xs), cons(log(last(xs)), ys))

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

last#(cons(x, cons(y, xs)))last#(cons(y, xs))

Rewrite Rules

le(s(x), 0)falsele(0, y)true
le(s(x), s(y))le(x, y)double(0)0
double(s(x))s(s(double(x)))log(0)logError
log(s(x))loop(s(x), s(0), 0)loop(x, s(y), z)if(le(x, s(y)), x, s(y), z)
if(true, x, y, z)zif(false, x, y, z)loop(x, double(y), s(z))
maplog(xs)mapIter(xs, nil)mapIter(xs, ys)ifmap(isempty(xs), xs, ys)
ifmap(true, xs, ys)ysifmap(false, xs, ys)mapIter(droplast(xs), cons(log(last(xs)), ys))
isempty(nil)trueisempty(cons(x, xs))false
last(nil)errorlast(cons(x, nil))x
last(cons(x, cons(y, xs)))last(cons(y, xs))droplast(nil)nil
droplast(cons(x, nil))nildroplast(cons(x, cons(y, xs)))cons(x, droplast(cons(y, xs)))
abac

Original Signature

Termination of terms over the following signature is verified: b, error, last, c, a, ifmap, true, logError, double, log, 0, mapIter, s, le, isempty, if, loop, false, droplast, maplog, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

last#(cons(x, cons(y, xs)))last#(cons(y, xs))

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

droplast#(cons(x, cons(y, xs)))droplast#(cons(y, xs))

Rewrite Rules

le(s(x), 0)falsele(0, y)true
le(s(x), s(y))le(x, y)double(0)0
double(s(x))s(s(double(x)))log(0)logError
log(s(x))loop(s(x), s(0), 0)loop(x, s(y), z)if(le(x, s(y)), x, s(y), z)
if(true, x, y, z)zif(false, x, y, z)loop(x, double(y), s(z))
maplog(xs)mapIter(xs, nil)mapIter(xs, ys)ifmap(isempty(xs), xs, ys)
ifmap(true, xs, ys)ysifmap(false, xs, ys)mapIter(droplast(xs), cons(log(last(xs)), ys))
isempty(nil)trueisempty(cons(x, xs))false
last(nil)errorlast(cons(x, nil))x
last(cons(x, cons(y, xs)))last(cons(y, xs))droplast(nil)nil
droplast(cons(x, nil))nildroplast(cons(x, cons(y, xs)))cons(x, droplast(cons(y, xs)))
abac

Original Signature

Termination of terms over the following signature is verified: b, error, last, c, a, ifmap, true, logError, double, log, 0, mapIter, s, le, isempty, if, loop, false, droplast, maplog, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

droplast#(cons(x, cons(y, xs)))droplast#(cons(y, xs))

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

le#(s(x), s(y))le#(x, y)

Rewrite Rules

le(s(x), 0)falsele(0, y)true
le(s(x), s(y))le(x, y)double(0)0
double(s(x))s(s(double(x)))log(0)logError
log(s(x))loop(s(x), s(0), 0)loop(x, s(y), z)if(le(x, s(y)), x, s(y), z)
if(true, x, y, z)zif(false, x, y, z)loop(x, double(y), s(z))
maplog(xs)mapIter(xs, nil)mapIter(xs, ys)ifmap(isempty(xs), xs, ys)
ifmap(true, xs, ys)ysifmap(false, xs, ys)mapIter(droplast(xs), cons(log(last(xs)), ys))
isempty(nil)trueisempty(cons(x, xs))false
last(nil)errorlast(cons(x, nil))x
last(cons(x, cons(y, xs)))last(cons(y, xs))droplast(nil)nil
droplast(cons(x, nil))nildroplast(cons(x, cons(y, xs)))cons(x, droplast(cons(y, xs)))
abac

Original Signature

Termination of terms over the following signature is verified: b, error, last, c, a, ifmap, true, logError, double, log, 0, mapIter, s, le, isempty, if, loop, false, droplast, maplog, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

le#(s(x), s(y))le#(x, y)

Problem 7: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

double#(s(x))double#(x)

Rewrite Rules

le(s(x), 0)falsele(0, y)true
le(s(x), s(y))le(x, y)double(0)0
double(s(x))s(s(double(x)))log(0)logError
log(s(x))loop(s(x), s(0), 0)loop(x, s(y), z)if(le(x, s(y)), x, s(y), z)
if(true, x, y, z)zif(false, x, y, z)loop(x, double(y), s(z))
maplog(xs)mapIter(xs, nil)mapIter(xs, ys)ifmap(isempty(xs), xs, ys)
ifmap(true, xs, ys)ysifmap(false, xs, ys)mapIter(droplast(xs), cons(log(last(xs)), ys))
isempty(nil)trueisempty(cons(x, xs))false
last(nil)errorlast(cons(x, nil))x
last(cons(x, cons(y, xs)))last(cons(y, xs))droplast(nil)nil
droplast(cons(x, nil))nildroplast(cons(x, cons(y, xs)))cons(x, droplast(cons(y, xs)))
abac

Original Signature

Termination of terms over the following signature is verified: b, error, last, c, a, ifmap, true, logError, double, log, 0, mapIter, s, le, isempty, if, loop, false, droplast, maplog, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

double#(s(x))double#(x)