YES

The TRS could be proven terminating. The proof took 545 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (12ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4iUR (478ms).
 |    | – Problem 4 was processed with processor DependencyGraph (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

rev2#(x, cons(y, l))rev2#(y, l)rev1#(x, cons(y, l))rev1#(y, l)
rev#(cons(x, l))rev2#(x, l)rev#(cons(x, l))rev1#(x, l)
rev2#(x, cons(y, l))rev#(cons(x, rev2(y, l)))

Rewrite Rules

rev(nil)nilrev(cons(x, l))cons(rev1(x, l), rev2(x, l))
rev1(0, nil)0rev1(s(x), nil)s(x)
rev1(x, cons(y, l))rev1(y, l)rev2(x, nil)nil
rev2(x, cons(y, l))rev(cons(x, rev2(y, l)))

Original Signature

Termination of terms over the following signature is verified: rev, rev1, rev2, 0, s, nil, cons

Strategy


The following SCCs where found

rev1#(x, cons(y, l)) → rev1#(y, l)

rev2#(x, cons(y, l)) → rev2#(y, l)rev#(cons(x, l)) → rev2#(x, l)
rev2#(x, cons(y, l)) → rev#(cons(x, rev2(y, l)))

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

rev1#(x, cons(y, l))rev1#(y, l)

Rewrite Rules

rev(nil)nilrev(cons(x, l))cons(rev1(x, l), rev2(x, l))
rev1(0, nil)0rev1(s(x), nil)s(x)
rev1(x, cons(y, l))rev1(y, l)rev2(x, nil)nil
rev2(x, cons(y, l))rev(cons(x, rev2(y, l)))

Original Signature

Termination of terms over the following signature is verified: rev, rev1, rev2, 0, s, nil, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

rev1#(x, cons(y, l))rev1#(y, l)

Problem 3: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

rev2#(x, cons(y, l))rev2#(y, l)rev#(cons(x, l))rev2#(x, l)
rev2#(x, cons(y, l))rev#(cons(x, rev2(y, l)))

Rewrite Rules

rev(nil)nilrev(cons(x, l))cons(rev1(x, l), rev2(x, l))
rev1(0, nil)0rev1(s(x), nil)s(x)
rev1(x, cons(y, l))rev1(y, l)rev2(x, nil)nil
rev2(x, cons(y, l))rev(cons(x, rev2(y, l)))

Original Signature

Termination of terms over the following signature is verified: rev, rev1, rev2, 0, s, nil, cons

Strategy


Polynomial Interpretation

Improved Usable rules

rev2(x, nil)nilrev(nil)nil
rev(cons(x, l))cons(rev1(x, l), rev2(x, l))rev2(x, cons(y, l))rev(cons(x, rev2(y, l)))

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

rev2#(x, cons(y, l))rev2#(y, l)rev#(cons(x, l))rev2#(x, l)

Problem 4: DependencyGraph



Dependency Pair Problem

Dependency Pairs

rev2#(x, cons(y, l))rev#(cons(x, rev2(y, l)))

Rewrite Rules

rev(nil)nilrev(cons(x, l))cons(rev1(x, l), rev2(x, l))
rev1(0, nil)0rev1(s(x), nil)s(x)
rev1(x, cons(y, l))rev1(y, l)rev2(x, nil)nil
rev2(x, cons(y, l))rev(cons(x, rev2(y, l)))

Original Signature

Termination of terms over the following signature is verified: rev1, rev, 0, rev2, s, cons, nil

Strategy


There are no SCCs!