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))xnot(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

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))