YES
The TRS could be proven terminating. The proof took 17 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (5ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(a, empty) | → | g#(a, empty) | | g#(cons(x, k), d) | → | g#(k, cons(x, d)) |
| f#(a, cons(x, k)) | → | f#(cons(x, a), k) |
Rewrite Rules
| f(a, empty) | → | g(a, empty) | | f(a, cons(x, k)) | → | f(cons(x, a), k) |
| g(empty, d) | → | d | | g(cons(x, k), d) | → | g(k, cons(x, d)) |
Original Signature
Termination of terms over the following signature is verified: f, g, empty, cons
Strategy
The following SCCs where found
| g#(cons(x, k), d) → g#(k, cons(x, d)) |
| f#(a, cons(x, k)) → f#(cons(x, a), k) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(a, cons(x, k)) | → | f#(cons(x, a), k) |
Rewrite Rules
| f(a, empty) | → | g(a, empty) | | f(a, cons(x, k)) | → | f(cons(x, a), k) |
| g(empty, d) | → | d | | g(cons(x, k), d) | → | g(k, cons(x, d)) |
Original Signature
Termination of terms over the following signature is verified: f, g, empty, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(a, cons(x, k)) | → | f#(cons(x, a), k) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| g#(cons(x, k), d) | → | g#(k, cons(x, d)) |
Rewrite Rules
| f(a, empty) | → | g(a, empty) | | f(a, cons(x, k)) | → | f(cons(x, a), k) |
| g(empty, d) | → | d | | g(cons(x, k), d) | → | g(k, cons(x, d)) |
Original Signature
Termination of terms over the following signature is verified: f, g, empty, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| g#(cons(x, k), d) | → | g#(k, cons(x, d)) |