YES

The TRS could be proven terminating. The proof took 890 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (114ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4 (138ms).
 |    | – Problem 3 was processed with processor PolynomialLinearRange4 (138ms).
 |    |    | – Problem 4 was processed with processor PolynomialLinearRange4 (152ms).
 |    |    |    | – Problem 5 was processed with processor PolynomialLinearRange4 (53ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

T(zip(XS, YS))zip#(XS, YS)T(zip(x_1, x_2))T(x_1)
T(oddNs)oddNs#T(incr(oddNs))incr#(oddNs)
T(repItems(XS))repItems#(XS)T(cons(x_1, x_2))T(x_2)
oddNs#incr#(pairNs)tail#(cons(X, XS))T(XS)
T(take(x_1, x_2))T(x_1)T(cons(x_1, x_2))T(x_1)
T(incr(XS))incr#(XS)T(zip(x_1, x_2))T(x_2)
T(repItems(x_1))T(x_1)T(take(x_1, x_2))T(x_2)
T(take(N, XS))take#(N, XS)T(incr(x_1))T(x_1)
oddNs#pairNs#

Rewrite Rules

pairNscons(0, incr(oddNs))oddNsincr(pairNs)
incr(cons(X, XS))cons(s(X), incr(XS))take(0, XS)nil
take(s(N), cons(X, XS))cons(X, take(N, XS))zip(nil, XS)nil
zip(X, nil)nilzip(cons(X, XS), cons(Y, YS))cons(pair(X, Y), zip(XS, YS))
tail(cons(X, XS))XSrepItems(nil)nil
repItems(cons(X, XS))cons(X, cons(X, repItems(XS)))

Original Signature

Termination of terms over the following signature is verified: zip, 0, pairNs, s, pair, take, repItems, incr, oddNs, tail, cons, nil

Strategy

Context-sensitive strategy:
μ(oddNs#) = μ(T) = μ(pairNs#) = μ(0) = μ(pairNs) = μ(oddNs) = μ(nil) = ∅
μ(incr#) = μ(tail#) = μ(tail) = μ(s) = μ(repItems#) = μ(repItems) = μ(incr) = μ(cons) = {1}
μ(zip) = μ(pair) = μ(zip#) = μ(take#) = μ(take) = {1, 2}


The following SCCs where found

T(take(x_1, x_2)) → T(x_2)T(zip(x_1, x_2)) → T(x_1)
T(incr(x_1)) → T(x_1)T(cons(x_1, x_2)) → T(x_2)
T(take(x_1, x_2)) → T(x_1)T(cons(x_1, x_2)) → T(x_1)
T(zip(x_1, x_2)) → T(x_2)T(repItems(x_1)) → T(x_1)

Problem 2: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(take(x_1, x_2))T(x_2)T(zip(x_1, x_2))T(x_1)
T(incr(x_1))T(x_1)T(cons(x_1, x_2))T(x_2)
T(take(x_1, x_2))T(x_1)T(cons(x_1, x_2))T(x_1)
T(zip(x_1, x_2))T(x_2)T(repItems(x_1))T(x_1)

Rewrite Rules

pairNscons(0, incr(oddNs))oddNsincr(pairNs)
incr(cons(X, XS))cons(s(X), incr(XS))take(0, XS)nil
take(s(N), cons(X, XS))cons(X, take(N, XS))zip(nil, XS)nil
zip(X, nil)nilzip(cons(X, XS), cons(Y, YS))cons(pair(X, Y), zip(XS, YS))
tail(cons(X, XS))XSrepItems(nil)nil
repItems(cons(X, XS))cons(X, cons(X, repItems(XS)))

Original Signature

Termination of terms over the following signature is verified: zip, 0, pairNs, s, pair, take, repItems, incr, oddNs, tail, cons, nil

Strategy

Context-sensitive strategy:
μ(oddNs#) = μ(T) = μ(pairNs#) = μ(0) = μ(pairNs) = μ(oddNs) = μ(nil) = ∅
μ(incr#) = μ(tail#) = μ(tail) = μ(s) = μ(repItems) = μ(repItems#) = μ(incr) = μ(cons) = {1}
μ(zip) = μ(pair) = μ(zip#) = μ(take#) = μ(take) = {1, 2}


Polynomial Interpretation

There are no usable rules

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

T(cons(x_1, x_2))T(x_2)T(cons(x_1, x_2))T(x_1)
T(repItems(x_1))T(x_1)

Problem 3: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(zip(x_1, x_2))T(x_1)T(take(x_1, x_2))T(x_2)
T(incr(x_1))T(x_1)T(take(x_1, x_2))T(x_1)
T(zip(x_1, x_2))T(x_2)

Rewrite Rules

pairNscons(0, incr(oddNs))oddNsincr(pairNs)
incr(cons(X, XS))cons(s(X), incr(XS))take(0, XS)nil
take(s(N), cons(X, XS))cons(X, take(N, XS))zip(nil, XS)nil
zip(X, nil)nilzip(cons(X, XS), cons(Y, YS))cons(pair(X, Y), zip(XS, YS))
tail(cons(X, XS))XSrepItems(nil)nil
repItems(cons(X, XS))cons(X, cons(X, repItems(XS)))

Original Signature

Termination of terms over the following signature is verified: zip, 0, s, pairNs, pair, repItems, take, incr, tail, oddNs, nil, cons

Strategy

Context-sensitive strategy:
μ(oddNs#) = μ(T) = μ(pairNs#) = μ(0) = μ(pairNs) = μ(oddNs) = μ(nil) = ∅
μ(incr#) = μ(tail#) = μ(tail) = μ(s) = μ(repItems#) = μ(repItems) = μ(incr) = μ(cons) = {1}
μ(zip) = μ(pair) = μ(zip#) = μ(take#) = μ(take) = {1, 2}


Polynomial Interpretation

There are no usable rules

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

T(take(x_1, x_2))T(x_2)T(take(x_1, x_2))T(x_1)

Problem 4: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(zip(x_1, x_2))T(x_1)T(incr(x_1))T(x_1)
T(zip(x_1, x_2))T(x_2)

Rewrite Rules

pairNscons(0, incr(oddNs))oddNsincr(pairNs)
incr(cons(X, XS))cons(s(X), incr(XS))take(0, XS)nil
take(s(N), cons(X, XS))cons(X, take(N, XS))zip(nil, XS)nil
zip(X, nil)nilzip(cons(X, XS), cons(Y, YS))cons(pair(X, Y), zip(XS, YS))
tail(cons(X, XS))XSrepItems(nil)nil
repItems(cons(X, XS))cons(X, cons(X, repItems(XS)))

Original Signature

Termination of terms over the following signature is verified: zip, 0, pairNs, s, pair, take, repItems, incr, oddNs, tail, cons, nil

Strategy

Context-sensitive strategy:
μ(oddNs#) = μ(T) = μ(pairNs#) = μ(0) = μ(pairNs) = μ(oddNs) = μ(nil) = ∅
μ(incr#) = μ(tail#) = μ(tail) = μ(s) = μ(repItems) = μ(repItems#) = μ(incr) = μ(cons) = {1}
μ(zip) = μ(pair) = μ(zip#) = μ(take#) = μ(take) = {1, 2}


Polynomial Interpretation

There are no usable rules

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

T(incr(x_1))T(x_1)

Problem 5: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

T(zip(x_1, x_2))T(x_1)T(zip(x_1, x_2))T(x_2)

Rewrite Rules

pairNscons(0, incr(oddNs))oddNsincr(pairNs)
incr(cons(X, XS))cons(s(X), incr(XS))take(0, XS)nil
take(s(N), cons(X, XS))cons(X, take(N, XS))zip(nil, XS)nil
zip(X, nil)nilzip(cons(X, XS), cons(Y, YS))cons(pair(X, Y), zip(XS, YS))
tail(cons(X, XS))XSrepItems(nil)nil
repItems(cons(X, XS))cons(X, cons(X, repItems(XS)))

Original Signature

Termination of terms over the following signature is verified: zip, 0, s, pairNs, pair, repItems, take, incr, tail, oddNs, nil, cons

Strategy

Context-sensitive strategy:
μ(oddNs#) = μ(T) = μ(pairNs#) = μ(0) = μ(pairNs) = μ(oddNs) = μ(nil) = ∅
μ(incr#) = μ(tail#) = μ(tail) = μ(s) = μ(repItems#) = μ(repItems) = μ(incr) = μ(cons) = {1}
μ(zip) = μ(pair) = μ(zip#) = μ(take#) = μ(take) = {1, 2}


Polynomial Interpretation

There are no usable rules

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

T(zip(x_1, x_2))T(x_1)T(zip(x_1, x_2))T(x_2)