YES

The TRS could be proven terminating. The proof took 445 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (15ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4 (92ms).
 |    | – Problem 3 was processed with processor PolynomialLinearRange4 (13ms).
 |    |    | – Problem 4 was processed with processor PolynomialLinearRange4 (76ms).

Problem 1: DependencyGraph

Dependency Pair Problem

Dependency Pairs

T(first(x_1, x_2)) → T(x_2)T(first(X, Z)) → first#(X, Z)
T(s(x_1)) → T(x_1)T(from(x_1)) → T(x_1)
T(first(x_1, x_2)) → T(x_1)T(from(s(X))) → from#(s(X))

Rewrite Rules

first(0, X) → nilfirst(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

Original Signature

Termination of terms over the following signature is verified: 0, s, from, first, nil, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = μ(nil) = ∅
μ(from#) = μ(s) = μ(from) = μ(cons) = {1}
μ(first#) = μ(first) = {1, 2}

The following SCCs where found

T(first(x_1, x_2)) → T(x_2)T(s(x_1)) → T(x_1)
T(from(x_1)) → T(x_1)T(first(x_1, x_2)) → T(x_1)

Problem 2: PolynomialLinearRange4

Dependency Pair Problem

Dependency Pairs

T(first(x_1, x_2)) → T(x_2)T(s(x_1)) → T(x_1)
T(from(x_1)) → T(x_1)T(first(x_1, x_2)) → T(x_1)

Rewrite Rules

first(0, X) → nilfirst(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

Original Signature

Termination of terms over the following signature is verified: 0, s, from, first, nil, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = μ(nil) = ∅
μ(s) = μ(from#) = μ(from) = μ(cons) = {1}
μ(first#) = μ(first) = {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(first(x_1, x_2)) → T(x_2)T(first(x_1, x_2)) → T(x_1)

Problem 3: PolynomialLinearRange4

Dependency Pair Problem

Dependency Pairs

T(s(x_1)) → T(x_1)T(from(x_1)) → T(x_1)

Rewrite Rules

first(0, X) → nilfirst(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

Original Signature

Termination of terms over the following signature is verified: 0, s, from, first, cons, nil

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = μ(nil) = ∅
μ(from#) = μ(s) = μ(from) = μ(cons) = {1}
μ(first#) = μ(first) = {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(from(x_1)) → T(x_1)

Problem 4: PolynomialLinearRange4

Dependency Pair Problem

Dependency Pairs

T(s(x_1)) → T(x_1)

Rewrite Rules

first(0, X) → nilfirst(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))

Original Signature

Termination of terms over the following signature is verified: 0, s, from, first, nil, cons

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = μ(nil) = ∅
μ(s) = μ(from#) = μ(from) = μ(cons) = {1}
μ(first#) = μ(first) = {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(s(x_1)) → T(x_1)