YES
The TRS could be proven terminating. The proof took 24 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (11ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| mem#(x, union(y, z)) | → | mem#(x, z) | | mem#(x, union(y, z)) | → | or#(mem(x, y), mem(x, z)) |
| mem#(x, union(y, z)) | → | mem#(x, y) |
Rewrite Rules
| or(true, y) | → | true | | or(x, true) | → | true |
| or(false, false) | → | false | | mem(x, nil) | → | false |
| mem(x, set(y)) | → | =(x, y) | | mem(x, union(y, z)) | → | or(mem(x, y), mem(x, z)) |
Original Signature
Termination of terms over the following signature is verified: or, set, union, mem, true, false, =, nil
Strategy
The following SCCs where found
| mem#(x, union(y, z)) → mem#(x, z) | mem#(x, union(y, z)) → mem#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| mem#(x, union(y, z)) | → | mem#(x, z) | | mem#(x, union(y, z)) | → | mem#(x, y) |
Rewrite Rules
| or(true, y) | → | true | | or(x, true) | → | true |
| or(false, false) | → | false | | mem(x, nil) | → | false |
| mem(x, set(y)) | → | =(x, y) | | mem(x, union(y, z)) | → | or(mem(x, y), mem(x, z)) |
Original Signature
Termination of terms over the following signature is verified: or, set, union, mem, true, false, =, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| mem#(x, union(y, z)) | → | mem#(x, z) | | mem#(x, union(y, z)) | → | mem#(x, y) |