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 (102ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (633ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (10432ms), DependencyGraph (1ms), ReductionPairSAT (800ms), DependencyGraph (1ms), SizeChangePrinciple (timeout)].
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 7 was processed with processor DependencyGraph (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 | – Problem 5 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 8 was processed with processor DependencyGraph (2ms).
 | – Problem 6 was processed with processor SubtermCriterion (0ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

minsort#(cons(x, y))minsort#(del(min(x, y), cons(x, y)))

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)if1(true, x, y, xs)min(x, xs)
if1(false, x, y, xs)min(y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))minsort(nil)nil
minsort(cons(x, y))cons(min(x, y), minsort(del(min(x, y), cons(x, y))))min(x, nil)x
min(x, cons(y, z))if1(le(x, y), x, y, z)del(x, nil)nil
del(x, cons(y, z))if2(eq(x, y), x, y, z)

Original Signature

Termination of terms over the following signature is verified: min, minsort, true, if1, if2, 0, le, s, false, del, eq, cons, nil


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

minsort#(cons(x, y))minsort#(del(min(x, y), cons(x, y)))del#(x, cons(y, z))if2#(eq(x, y), x, y, z)
le#(s(x), s(y))le#(x, y)minsort#(cons(x, y))min#(x, y)
del#(x, cons(y, z))eq#(x, y)eq#(s(x), s(y))eq#(x, y)
if1#(false, x, y, xs)min#(y, xs)if2#(false, x, y, xs)del#(x, xs)
if1#(true, x, y, xs)min#(x, xs)min#(x, cons(y, z))if1#(le(x, y), x, y, z)
min#(x, cons(y, z))le#(x, y)minsort#(cons(x, y))del#(min(x, y), cons(x, y))

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)if1(true, x, y, xs)min(x, xs)
if1(false, x, y, xs)min(y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))minsort(nil)nil
minsort(cons(x, y))cons(min(x, y), minsort(del(min(x, y), cons(x, y))))min(x, nil)x
min(x, cons(y, z))if1(le(x, y), x, y, z)del(x, nil)nil
del(x, cons(y, z))if2(eq(x, y), x, y, z)

Original Signature

Termination of terms over the following signature is verified: min, minsort, true, if1, if2, 0, le, s, false, del, eq, cons, nil

Strategy


The following SCCs where found

minsort#(cons(x, y)) → minsort#(del(min(x, y), cons(x, y)))

le#(s(x), s(y)) → le#(x, y)

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

eq#(s(x), s(y)) → eq#(x, y)

if1#(false, x, y, xs) → min#(y, xs)if1#(true, x, y, xs) → min#(x, xs)
min#(x, cons(y, z)) → if1#(le(x, y), x, y, z)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)if1(true, x, y, xs)min(x, xs)
if1(false, x, y, xs)min(y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))minsort(nil)nil
minsort(cons(x, y))cons(min(x, y), minsort(del(min(x, y), cons(x, y))))min(x, nil)x
min(x, cons(y, z))if1(le(x, y), x, y, z)del(x, nil)nil
del(x, cons(y, z))if2(eq(x, y), x, y, z)

Original Signature

Termination of terms over the following signature is verified: min, minsort, true, if1, if2, 0, le, s, false, del, eq, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 7: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)if1(true, x, y, xs)min(x, xs)
if1(false, x, y, xs)min(y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))minsort(nil)nil
minsort(cons(x, y))cons(min(x, y), minsort(del(min(x, y), cons(x, y))))min(x, nil)x
min(x, cons(y, z))if1(le(x, y), x, y, z)del(x, nil)nil
del(x, cons(y, z))if2(eq(x, y), x, y, z)

Original Signature

Termination of terms over the following signature is verified: min, minsort, true, if1, if2, 0, le, s, false, del, eq, cons, nil

Strategy


There are no SCCs!

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)if1(true, x, y, xs)min(x, xs)
if1(false, x, y, xs)min(y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))minsort(nil)nil
minsort(cons(x, y))cons(min(x, y), minsort(del(min(x, y), cons(x, y))))min(x, nil)x
min(x, cons(y, z))if1(le(x, y), x, y, z)del(x, nil)nil
del(x, cons(y, z))if2(eq(x, y), x, y, z)

Original Signature

Termination of terms over the following signature is verified: min, minsort, true, if1, if2, 0, le, s, false, del, eq, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

if1#(false, x, y, xs)min#(y, xs)if1#(true, x, y, xs)min#(x, xs)
min#(x, cons(y, z))if1#(le(x, y), x, y, z)

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)if1(true, x, y, xs)min(x, xs)
if1(false, x, y, xs)min(y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))minsort(nil)nil
minsort(cons(x, y))cons(min(x, y), minsort(del(min(x, y), cons(x, y))))min(x, nil)x
min(x, cons(y, z))if1(le(x, y), x, y, z)del(x, nil)nil
del(x, cons(y, z))if2(eq(x, y), x, y, z)

Original Signature

Termination of terms over the following signature is verified: min, minsort, true, if1, if2, 0, le, s, false, del, eq, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

min#(x, cons(y, z))if1#(le(x, y), x, y, z)

Problem 8: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if1#(false, x, y, xs)min#(y, xs)if1#(true, x, y, xs)min#(x, xs)

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)if1(true, x, y, xs)min(x, xs)
if1(false, x, y, xs)min(y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))minsort(nil)nil
minsort(cons(x, y))cons(min(x, y), minsort(del(min(x, y), cons(x, y))))min(x, nil)x
min(x, cons(y, z))if1(le(x, y), x, y, z)del(x, nil)nil
del(x, cons(y, z))if2(eq(x, y), x, y, z)

Original Signature

Termination of terms over the following signature is verified: min, minsort, true, if1, if2, 0, le, s, false, del, eq, cons, nil

Strategy


There are no SCCs!

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)eq(0, 0)true
eq(0, s(y))falseeq(s(x), 0)false
eq(s(x), s(y))eq(x, y)if1(true, x, y, xs)min(x, xs)
if1(false, x, y, xs)min(y, xs)if2(true, x, y, xs)xs
if2(false, x, y, xs)cons(y, del(x, xs))minsort(nil)nil
minsort(cons(x, y))cons(min(x, y), minsort(del(min(x, y), cons(x, y))))min(x, nil)x
min(x, cons(y, z))if1(le(x, y), x, y, z)del(x, nil)nil
del(x, cons(y, z))if2(eq(x, y), x, y, z)

Original Signature

Termination of terms over the following signature is verified: min, minsort, true, if1, if2, 0, le, s, false, del, eq, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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