TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60001 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (106ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (920ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (753ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (9996ms), DependencyGraph (1ms), ReductionPairSAT (831ms), DependencyGraph (1ms), SizeChangePrinciple (timeout)].
 | – Problem 3 was processed with processor PolynomialLinearRange4iUR (221ms).
 |    | – Problem 8 was processed with processor DependencyGraph (2ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 | – Problem 5 was processed with processor SubtermCriterion (2ms).
 |    | – Problem 7 was processed with processor DependencyGraph (0ms).
 | – Problem 6 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

sort#(cons(x, xs))sort#(del(max(cons(x, xs)), cons(x, xs)))

Rewrite Rules

max(nil)0max(cons(x, nil))x
max(cons(x, cons(y, xs)))if1(ge(x, y), x, y, xs)if1(true, x, y, xs)max(cons(x, xs))
if1(false, x, y, xs)max(cons(y, xs))del(x, nil)nil
del(x, cons(y, xs))if2(eq(x, y), x, y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)sort(nil)nil
sort(cons(x, xs))cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))ge(0, 0)true
ge(s(x), 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: max, sort, true, if1, ge, if2, 0, s, false, del, nil, cons, eq


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if1#(false, x, y, xs)max#(cons(y, xs))sort#(cons(x, xs))del#(max(cons(x, xs)), cons(x, xs))
if1#(true, x, y, xs)max#(cons(x, xs))sort#(cons(x, xs))max#(cons(x, xs))
max#(cons(x, cons(y, xs)))ge#(x, y)del#(x, cons(y, xs))eq#(x, y)
del#(x, cons(y, xs))if2#(eq(x, y), x, y, xs)ge#(s(x), s(y))ge#(x, y)
eq#(s(x), s(y))eq#(x, y)max#(cons(x, cons(y, xs)))if1#(ge(x, y), x, y, xs)
if2#(false, x, y, xs)del#(x, xs)sort#(cons(x, xs))sort#(del(max(cons(x, xs)), cons(x, xs)))

Rewrite Rules

max(nil)0max(cons(x, nil))x
max(cons(x, cons(y, xs)))if1(ge(x, y), x, y, xs)if1(true, x, y, xs)max(cons(x, xs))
if1(false, x, y, xs)max(cons(y, xs))del(x, nil)nil
del(x, cons(y, xs))if2(eq(x, y), x, y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)sort(nil)nil
sort(cons(x, xs))cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))ge(0, 0)true
ge(s(x), 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: max, sort, true, if1, ge, if2, 0, s, false, del, nil, cons, eq

Strategy


The following SCCs where found

ge#(s(x), s(y)) → ge#(x, y)

del#(x, cons(y, xs)) → if2#(eq(x, y), x, y, xs)if2#(false, x, y, xs) → del#(x, xs)

eq#(s(x), s(y)) → eq#(x, y)

if1#(false, x, y, xs) → max#(cons(y, xs))if1#(true, x, y, xs) → max#(cons(x, xs))
max#(cons(x, cons(y, xs))) → if1#(ge(x, y), x, y, xs)

sort#(cons(x, xs)) → sort#(del(max(cons(x, xs)), cons(x, xs)))

Problem 3: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

if1#(false, x, y, xs)max#(cons(y, xs))if1#(true, x, y, xs)max#(cons(x, xs))
max#(cons(x, cons(y, xs)))if1#(ge(x, y), x, y, xs)

Rewrite Rules

max(nil)0max(cons(x, nil))x
max(cons(x, cons(y, xs)))if1(ge(x, y), x, y, xs)if1(true, x, y, xs)max(cons(x, xs))
if1(false, x, y, xs)max(cons(y, xs))del(x, nil)nil
del(x, cons(y, xs))if2(eq(x, y), x, y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)sort(nil)nil
sort(cons(x, xs))cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))ge(0, 0)true
ge(s(x), 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: max, sort, true, if1, ge, if2, 0, s, false, del, nil, cons, eq

Strategy


Polynomial Interpretation

Improved Usable rules

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

max#(cons(x, cons(y, xs)))if1#(ge(x, y), x, y, xs)

Problem 8: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if1#(false, x, y, xs)max#(cons(y, xs))if1#(true, x, y, xs)max#(cons(x, xs))

Rewrite Rules

max(nil)0max(cons(x, nil))x
max(cons(x, cons(y, xs)))if1(ge(x, y), x, y, xs)if1(true, x, y, xs)max(cons(x, xs))
if1(false, x, y, xs)max(cons(y, xs))del(x, nil)nil
del(x, cons(y, xs))if2(eq(x, y), x, y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)sort(nil)nil
sort(cons(x, xs))cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))ge(0, 0)true
ge(s(x), 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: max, sort, true, if1, ge, if2, 0, s, false, del, nil, cons, eq

Strategy


There are no SCCs!

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

max(nil)0max(cons(x, nil))x
max(cons(x, cons(y, xs)))if1(ge(x, y), x, y, xs)if1(true, x, y, xs)max(cons(x, xs))
if1(false, x, y, xs)max(cons(y, xs))del(x, nil)nil
del(x, cons(y, xs))if2(eq(x, y), x, y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)sort(nil)nil
sort(cons(x, xs))cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))ge(0, 0)true
ge(s(x), 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: max, sort, true, if1, ge, if2, 0, s, false, del, nil, cons, eq

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

del#(x, cons(y, xs))if2#(eq(x, y), x, y, xs)if2#(false, x, y, xs)del#(x, xs)

Rewrite Rules

max(nil)0max(cons(x, nil))x
max(cons(x, cons(y, xs)))if1(ge(x, y), x, y, xs)if1(true, x, y, xs)max(cons(x, xs))
if1(false, x, y, xs)max(cons(y, xs))del(x, nil)nil
del(x, cons(y, xs))if2(eq(x, y), x, y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)sort(nil)nil
sort(cons(x, xs))cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))ge(0, 0)true
ge(s(x), 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: max, sort, true, if1, ge, if2, 0, s, false, del, nil, cons, eq

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

del#(x, cons(y, xs))if2#(eq(x, y), x, y, xs)

Problem 7: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if2#(false, x, y, xs)del#(x, xs)

Rewrite Rules

max(nil)0max(cons(x, nil))x
max(cons(x, cons(y, xs)))if1(ge(x, y), x, y, xs)if1(true, x, y, xs)max(cons(x, xs))
if1(false, x, y, xs)max(cons(y, xs))del(x, nil)nil
del(x, cons(y, xs))if2(eq(x, y), x, y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)sort(nil)nil
sort(cons(x, xs))cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))ge(0, 0)true
ge(s(x), 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: max, sort, true, if1, ge, if2, 0, s, false, del, nil, cons, eq

Strategy


There are no SCCs!

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

ge#(s(x), s(y))ge#(x, y)

Rewrite Rules

max(nil)0max(cons(x, nil))x
max(cons(x, cons(y, xs)))if1(ge(x, y), x, y, xs)if1(true, x, y, xs)max(cons(x, xs))
if1(false, x, y, xs)max(cons(y, xs))del(x, nil)nil
del(x, cons(y, xs))if2(eq(x, y), x, y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)sort(nil)nil
sort(cons(x, xs))cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs))))ge(0, 0)true
ge(s(x), 0)truege(0, s(x))false
ge(s(x), s(y))ge(x, y)

Original Signature

Termination of terms over the following signature is verified: max, sort, true, if1, ge, if2, 0, s, false, del, nil, cons, eq

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

ge#(s(x), s(y))ge#(x, y)