MAYBE

The TRS could not be proven terminating. The proof attempt took 279 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (0ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (67ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (31ms), DependencyGraph (1ms), ReductionPairSAT (24ms), DependencyGraph (1ms), SizeChangePrinciple (4ms)].
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor SubtermCriterion (0ms).
 | – Problem 5 was processed with processor SubtermCriterion (0ms).
 | – Problem 6 was processed with processor SubtermCriterion (0ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

terms#(N)terms#(s(N))

Rewrite Rules

terms(N)cons(recip(sqr(N)), terms(s(N)))sqr(0)0
sqr(s(X))s(add(sqr(X), dbl(X)))dbl(0)0
dbl(s(X))s(s(dbl(X)))add(0, X)X
add(s(X), Y)s(add(X, Y))first(0, X)nil
first(s(X), cons(Y, Z))cons(Y, first(X, Z))

Original Signature

Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, nil, cons


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

sqr#(s(X))add#(sqr(X), dbl(X))add#(s(X), Y)add#(X, Y)
first#(s(X), cons(Y, Z))first#(X, Z)dbl#(s(X))dbl#(X)
terms#(N)sqr#(N)terms#(N)terms#(s(N))
sqr#(s(X))dbl#(X)sqr#(s(X))sqr#(X)

Rewrite Rules

terms(N)cons(recip(sqr(N)), terms(s(N)))sqr(0)0
sqr(s(X))s(add(sqr(X), dbl(X)))dbl(0)0
dbl(s(X))s(s(dbl(X)))add(0, X)X
add(s(X), Y)s(add(X, Y))first(0, X)nil
first(s(X), cons(Y, Z))cons(Y, first(X, Z))

Original Signature

Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil

Strategy


The following SCCs where found

add#(s(X), Y) → add#(X, Y)

first#(s(X), cons(Y, Z)) → first#(X, Z)

dbl#(s(X)) → dbl#(X)

terms#(N) → terms#(s(N))

sqr#(s(X)) → sqr#(X)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

sqr#(s(X))sqr#(X)

Rewrite Rules

terms(N)cons(recip(sqr(N)), terms(s(N)))sqr(0)0
sqr(s(X))s(add(sqr(X), dbl(X)))dbl(0)0
dbl(s(X))s(s(dbl(X)))add(0, X)X
add(s(X), Y)s(add(X, Y))first(0, X)nil
first(s(X), cons(Y, Z))cons(Y, first(X, Z))

Original Signature

Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

sqr#(s(X))sqr#(X)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

dbl#(s(X))dbl#(X)

Rewrite Rules

terms(N)cons(recip(sqr(N)), terms(s(N)))sqr(0)0
sqr(s(X))s(add(sqr(X), dbl(X)))dbl(0)0
dbl(s(X))s(s(dbl(X)))add(0, X)X
add(s(X), Y)s(add(X, Y))first(0, X)nil
first(s(X), cons(Y, Z))cons(Y, first(X, Z))

Original Signature

Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

dbl#(s(X))dbl#(X)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

first#(s(X), cons(Y, Z))first#(X, Z)

Rewrite Rules

terms(N)cons(recip(sqr(N)), terms(s(N)))sqr(0)0
sqr(s(X))s(add(sqr(X), dbl(X)))dbl(0)0
dbl(s(X))s(s(dbl(X)))add(0, X)X
add(s(X), Y)s(add(X, Y))first(0, X)nil
first(s(X), cons(Y, Z))cons(Y, first(X, Z))

Original Signature

Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

first#(s(X), cons(Y, Z))first#(X, Z)

Problem 6: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

add#(s(X), Y)add#(X, Y)

Rewrite Rules

terms(N)cons(recip(sqr(N)), terms(s(N)))sqr(0)0
sqr(s(X))s(add(sqr(X), dbl(X)))dbl(0)0
dbl(s(X))s(s(dbl(X)))add(0, X)X
add(s(X), Y)s(add(X, Y))first(0, X)nil
first(s(X), cons(Y, Z))cons(Y, first(X, Z))

Original Signature

Termination of terms over the following signature is verified: 0, s, terms, sqr, dbl, recip, add, first, cons, nil

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

add#(s(X), Y)add#(X, Y)