The rewrite relation of the following TRS is considered.
| purge(nil) | → | nil | (1) |
| purge(.(x,y)) | → | .(x,purge(remove(x,y))) | (2) |
| remove(x,nil) | → | nil | (3) |
| remove(x,.(y,z)) | → | if(=(x,y),remove(x,z),.(y,remove(x,z))) | (4) |
| purge#(.(x,y)) | → | remove#(x,y) | (5) |
| purge#(.(x,y)) | → | purge#(remove(x,y)) | (6) |
| remove#(x,.(y,z)) | → | remove#(x,z) | (7) |
The dependency pairs are split into 1 component.
| remove#(x,.(y,z)) | → | remove#(x,z) | (7) |
Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.
| remove#(x,.(y,z)) | → | remove#(x,z) | (7) |
| 2 | > | 2 | |
| 1 | ≥ | 1 |
As there is no critical graph in the transitive closure, there are no infinite chains.