YES
The TRS could be proven terminating. The proof took 1273 ms.
The following DP Processors were used
Problem 1 was processed with processor PolynomialLinearRange4iUR (898ms).
Problem 1: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| not#(and(x, y)) | → | not#(not(not(y))) | | not#(and(x, y)) | → | not#(x) |
| not#(or(x, y)) | → | not#(not(not(x))) | | not#(and(x, y)) | → | not#(not(not(x))) |
| not#(and(x, y)) | → | not#(not(y)) | | not#(or(x, y)) | → | not#(x) |
| not#(or(x, y)) | → | not#(not(not(y))) | | not#(or(x, y)) | → | not#(not(x)) |
| not#(and(x, y)) | → | not#(not(x)) | | not#(and(x, y)) | → | not#(y) |
| not#(or(x, y)) | → | not#(y) | | not#(or(x, y)) | → | not#(not(y)) |
Rewrite Rules
| not(not(x)) | → | x | | not(or(x, y)) | → | and(not(not(not(x))), not(not(not(y)))) |
| not(and(x, y)) | → | or(not(not(not(x))), not(not(not(y)))) |
Original Signature
Termination of terms over the following signature is verified: not, or, and
Strategy
Polynomial Interpretation
- and(x,y): 2y + x + 1
- not(x): x
- not#(x): 2x
- or(x,y): 2y + x + 1
Improved Usable rules
| not(or(x, y)) | → | and(not(not(not(x))), not(not(not(y)))) | | not(and(x, y)) | → | or(not(not(not(x))), not(not(not(y)))) |
| not(not(x)) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| not#(and(x, y)) | → | not#(not(not(y))) | | not#(and(x, y)) | → | not#(x) |
| not#(or(x, y)) | → | not#(not(not(x))) | | not#(and(x, y)) | → | not#(not(not(x))) |
| not#(and(x, y)) | → | not#(not(y)) | | not#(or(x, y)) | → | not#(not(not(y))) |
| not#(or(x, y)) | → | not#(x) | | not#(and(x, y)) | → | not#(not(x)) |
| not#(or(x, y)) | → | not#(not(x)) | | not#(and(x, y)) | → | not#(y) |
| not#(or(x, y)) | → | not#(y) | | not#(or(x, y)) | → | not#(not(y)) |