YES

The TRS could be proven terminating. The proof took 839 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (6ms).
 | – Problem 2 was processed with processor PolynomialOrderingProcessor (208ms).
 |    | – Problem 3 was processed with processor PolynomialOrderingProcessor (271ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy


The following SCCs where found

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

Problem 2: PolynomialOrderingProcessor



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy


Polynomial Interpretation

Improved Usable rules

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

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

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

Problem 3: PolynomialOrderingProcessor



Dependency Pair Problem

Dependency Pairs

f#(f(a, x), a) → f#(f(a, f(a, a)), x)

Rewrite Rules

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

Original Signature

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

Strategy


Polynomial Interpretation

Improved Usable rules

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

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

f#(f(a, x), a) → f#(f(a, f(a, a)), x)