YES
The TRS could be proven terminating. The proof took 37 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (4ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| | – Problem 3 was processed with processor DependencyGraph (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| evenodd#(x, 0) | → | not#(evenodd(x, s(0))) | | evenodd#(s(x), s(0)) | → | evenodd#(x, 0) |
| evenodd#(x, 0) | → | evenodd#(x, s(0)) |
Rewrite Rules
| not(true) | → | false | | not(false) | → | true |
| evenodd(x, 0) | → | not(evenodd(x, s(0))) | | evenodd(0, s(0)) | → | false |
| evenodd(s(x), s(0)) | → | evenodd(x, 0) |
Original Signature
Termination of terms over the following signature is verified: not, 0, s, false, true, evenodd
Strategy
The following SCCs where found
| evenodd#(s(x), s(0)) → evenodd#(x, 0) | evenodd#(x, 0) → evenodd#(x, s(0)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| evenodd#(s(x), s(0)) | → | evenodd#(x, 0) | | evenodd#(x, 0) | → | evenodd#(x, s(0)) |
Rewrite Rules
| not(true) | → | false | | not(false) | → | true |
| evenodd(x, 0) | → | not(evenodd(x, s(0))) | | evenodd(0, s(0)) | → | false |
| evenodd(s(x), s(0)) | → | evenodd(x, 0) |
Original Signature
Termination of terms over the following signature is verified: not, 0, s, false, true, evenodd
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| evenodd#(s(x), s(0)) | → | evenodd#(x, 0) |
Problem 3: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| evenodd#(x, 0) | → | evenodd#(x, s(0)) |
Rewrite Rules
| not(true) | → | false | | not(false) | → | true |
| evenodd(x, 0) | → | not(evenodd(x, s(0))) | | evenodd(0, s(0)) | → | false |
| evenodd(s(x), s(0)) | → | evenodd(x, 0) |
Original Signature
Termination of terms over the following signature is verified: not, 0, s, true, false, evenodd
Strategy
There are no SCCs!