YES
The TRS could be proven terminating. The proof took 159 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (3ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (138ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(c(X, s(Y))) | → | f#(c(s(X), Y)) | | g#(c(s(X), Y)) | → | f#(c(X, s(Y))) |
Rewrite Rules
| f(c(X, s(Y))) | → | f(c(s(X), Y)) | | g(c(s(X), Y)) | → | f(c(X, s(Y))) |
Original Signature
Termination of terms over the following signature is verified: f, g, s, c
Strategy
The following SCCs where found
| f#(c(X, s(Y))) → f#(c(s(X), Y)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(c(X, s(Y))) | → | f#(c(s(X), Y)) |
Rewrite Rules
| f(c(X, s(Y))) | → | f(c(s(X), Y)) | | g(c(s(X), Y)) | → | f(c(X, s(Y))) |
Original Signature
Termination of terms over the following signature is verified: f, g, s, c
Strategy
Polynomial Interpretation
- c(x,y): 2y
- f(x): 0
- f#(x): x
- g(x): 0
- s(x): 2x + 1
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(c(X, s(Y))) | → | f#(c(s(X), Y)) |