YES

The TRS could be proven terminating. The proof took 29 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (16ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

g#(s(X))g#(X)sel#(s(X), cons(Y, Z))activate#(Z)
f#(X)g#(X)activate#(n__f(X))f#(X)
sel#(s(X), cons(Y, Z))sel#(X, activate(Z))

Rewrite Rules

f(X)cons(X, n__f(g(X)))g(0)s(0)
g(s(X))s(s(g(X)))sel(0, cons(X, Y))X
sel(s(X), cons(Y, Z))sel(X, activate(Z))f(X)n__f(X)
activate(n__f(X))f(X)activate(X)X

Original Signature

Termination of terms over the following signature is verified: f, activate, g, 0, s, sel, n__f, cons

Strategy


The following SCCs where found

g#(s(X)) → g#(X)

sel#(s(X), cons(Y, Z)) → sel#(X, activate(Z))

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

sel#(s(X), cons(Y, Z))sel#(X, activate(Z))

Rewrite Rules

f(X)cons(X, n__f(g(X)))g(0)s(0)
g(s(X))s(s(g(X)))sel(0, cons(X, Y))X
sel(s(X), cons(Y, Z))sel(X, activate(Z))f(X)n__f(X)
activate(n__f(X))f(X)activate(X)X

Original Signature

Termination of terms over the following signature is verified: f, activate, g, 0, s, sel, n__f, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

sel#(s(X), cons(Y, Z))sel#(X, activate(Z))

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

g#(s(X))g#(X)

Rewrite Rules

f(X)cons(X, n__f(g(X)))g(0)s(0)
g(s(X))s(s(g(X)))sel(0, cons(X, Y))X
sel(s(X), cons(Y, Z))sel(X, activate(Z))f(X)n__f(X)
activate(n__f(X))f(X)activate(X)X

Original Signature

Termination of terms over the following signature is verified: f, activate, g, 0, s, sel, n__f, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

g#(s(X))g#(X)