YES
The TRS could be proven terminating. The proof took 243 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (4ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (173ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(s(x), s(y)) | → | f#(x, y) | | g#(0, x) | → | g#(f(x, x), x) |
| g#(0, x) | → | f#(x, x) |
Rewrite Rules
| f(x, 0) | → | s(0) | | f(s(x), s(y)) | → | s(f(x, y)) |
| g(0, x) | → | g(f(x, x), x) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s
Strategy
The following SCCs where found
| f#(s(x), s(y)) → f#(x, y) |
| g#(0, x) → g#(f(x, x), x) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| g#(0, x) | → | g#(f(x, x), x) |
Rewrite Rules
| f(x, 0) | → | s(0) | | f(s(x), s(y)) | → | s(f(x, y)) |
| g(0, x) | → | g(f(x, x), x) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s
Strategy
Polynomial Interpretation
- 0: 1
- f(x,y): 0
- g(x,y): 0
- g#(x,y): 2x + 1
- s(x): 0
Improved Usable rules
| f(x, 0) | → | s(0) | | f(s(x), s(y)) | → | s(f(x, y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| g#(0, x) | → | g#(f(x, x), x) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(s(x), s(y)) | → | f#(x, y) |
Rewrite Rules
| f(x, 0) | → | s(0) | | f(s(x), s(y)) | → | s(f(x, y)) |
| g(0, x) | → | g(f(x, x), x) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(s(x), s(y)) | → | f#(x, y) |