YES
The TRS could be proven terminating. The proof took 53 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (16ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| not#(and(x, y)) | → | not#(x) | | not#(or(x, y)) | → | and#(not(x), not(y)) |
| not#(or(x, y)) | → | not#(x) | | not#(and(x, y)) | → | not#(y) |
| not#(or(x, y)) | → | not#(y) | | not#(and(x, y)) | → | or#(not(x), not(y)) |
Rewrite Rules
| or(x, x) | → | x | | and(x, x) | → | x |
| not(not(x)) | → | x | | not(and(x, y)) | → | or(not(x), not(y)) |
| not(or(x, y)) | → | and(not(x), not(y)) |
Original Signature
Termination of terms over the following signature is verified: not, or, and
Strategy
The following SCCs where found
| not#(and(x, y)) → not#(x) | not#(or(x, y)) → not#(x) |
| not#(and(x, y)) → not#(y) | not#(or(x, y)) → not#(y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| not#(and(x, y)) | → | not#(x) | | not#(or(x, y)) | → | not#(x) |
| not#(and(x, y)) | → | not#(y) | | not#(or(x, y)) | → | not#(y) |
Rewrite Rules
| or(x, x) | → | x | | and(x, x) | → | x |
| not(not(x)) | → | x | | not(and(x, y)) | → | or(not(x), not(y)) |
| not(or(x, y)) | → | and(not(x), not(y)) |
Original Signature
Termination of terms over the following signature is verified: not, or, and
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| not#(and(x, y)) | → | not#(x) | | not#(or(x, y)) | → | not#(x) |
| not#(and(x, y)) | → | not#(y) | | not#(or(x, y)) | → | not#(y) |