YES

The TRS could be proven terminating. The proof took 60000 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (5ms).
 | – Problem 2 was processed with processor PolynomialOrderingProcessor (382ms).
 |    | – Problem 3 was processed with processor PolynomialOrderingProcessor (213ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

f#(a, f(x, a)) → f#(a, a)f#(a, f(x, a)) → f#(f(a, a), a)
f#(a, f(x, a)) → f#(a, f(x, f(f(a, a), a)))f#(a, f(x, a)) → f#(x, f(f(a, a), a))

Rewrite Rules

f(a, f(x, a)) → f(a, f(x, f(f(a, a), a)))

Original Signature

Termination of terms over the following signature is verified: f, a

Strategy


The following SCCs where found

f#(a, f(x, a)) → f#(a, f(x, f(f(a, a), a)))f#(a, f(x, a)) → f#(x, f(f(a, a), a))

Problem 2: PolynomialOrderingProcessor



Dependency Pair Problem

Dependency Pairs

f#(a, f(x, a)) → f#(a, f(x, f(f(a, a), a)))f#(a, f(x, a)) → f#(x, f(f(a, a), a))

Rewrite Rules

f(a, f(x, a)) → f(a, f(x, f(f(a, a), a)))

Original Signature

Termination of terms over the following signature is verified: f, a

Strategy


Polynomial Interpretation

Improved Usable rules

f(a, f(x, a)) → f(a, f(x, f(f(a, a), a)))

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

f#(a, f(x, a)) → f#(a, f(x, f(f(a, a), a)))

Problem 3: PolynomialOrderingProcessor



Dependency Pair Problem

Dependency Pairs

f#(a, f(x, a)) → f#(x, f(f(a, a), a))

Rewrite Rules

f(a, f(x, a)) → f(a, f(x, f(f(a, a), a)))

Original Signature

Termination of terms over the following signature is verified: f, a

Strategy


Polynomial Interpretation

Improved Usable rules

f(a, f(x, a)) → f(a, f(x, f(f(a, a), a)))

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

f#(a, f(x, a)) → f#(x, f(f(a, a), a))