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)