YES

The TRS could be proven terminating. The proof took 465 ms.

The following DP Processors were used


Problem 1 was processed with processor PolynomialLinearRange4iUR (169ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4iUR (81ms).

Problem 1: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

f#(s(x), y)f#(f(x, y), y)f#(s(x), y)f#(x, y)

Rewrite Rules

f(0, y)0f(s(x), y)f(f(x, y), y)

Original Signature

Termination of terms over the following signature is verified: f, 0, s

Strategy


Polynomial Interpretation

Improved Usable rules

f(0, y)0f(s(x), y)f(f(x, y), y)

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

f#(s(x), y)f#(x, y)

Problem 2: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

f#(s(x), y)f#(f(x, y), y)

Rewrite Rules

f(0, y)0f(s(x), y)f(f(x, y), y)

Original Signature

Termination of terms over the following signature is verified: f, 0, s

Strategy


Polynomial Interpretation

Improved Usable rules

f(0, y)0f(s(x), y)f(f(x, y), y)

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

f#(s(x), y)f#(f(x, y), y)