TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60005 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (78ms).
 | – Problem 2 was processed with processor PolynomialLinearRange8NegiUR (6261ms).
 |    | – Problem 3 was processed with processor PolynomialLinearRange8NegiUR (11221ms).
 |    |    | – Problem 4 was processed with processor PolynomialLinearRange8NegiUR (15488ms).
 |    |    |    | – Problem 5 was processed with processor PolynomialLinearRange8NegiUR (2299ms).
 |    |    |    |    | – Problem 6 remains open; application of the following processors failed [DependencyGraph (1ms), PolynomialLinearRange8NegiUR (timeout)].

The following open problems remain:



Open Dependency Pair Problem 6

Dependency Pairs

b#(0, a(1, a(x, y)))a#(0, a(x, y))a#(0, b(0, x))b#(0, a(0, x))

Rewrite Rules

a(0, b(0, x))b(0, a(0, x))a(0, x)b(0, b(0, x))
a(0, a(1, a(x, y)))a(1, a(0, a(x, y)))b(0, a(1, a(x, y)))b(1, a(0, a(x, y)))
a(0, a(x, y))a(1, a(1, a(x, y)))

Original Signature

Termination of terms over the following signature is verified: 1, 0, b, a


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

b#(0, a(1, a(x, y)))a#(0, a(x, y))a#(0, b(0, x))b#(0, a(0, x))
a#(0, a(1, a(x, y)))a#(1, a(0, a(x, y)))b#(0, a(1, a(x, y)))b#(1, a(0, a(x, y)))
a#(0, a(x, y))a#(x, y)a#(0, b(0, x))a#(0, x)
a#(0, a(1, a(x, y)))a#(0, a(x, y))b#(0, a(1, a(x, y)))a#(x, y)
a#(0, a(1, a(x, y)))a#(x, y)a#(0, a(x, y))a#(1, a(1, a(x, y)))
a#(0, x)b#(0, b(0, x))a#(0, x)b#(0, x)
a#(0, a(x, y))a#(1, a(x, y))

Rewrite Rules

a(0, b(0, x))b(0, a(0, x))a(0, x)b(0, b(0, x))
a(0, a(1, a(x, y)))a(1, a(0, a(x, y)))b(0, a(1, a(x, y)))b(1, a(0, a(x, y)))
a(0, a(x, y))a(1, a(1, a(x, y)))

Original Signature

Termination of terms over the following signature is verified: 1, 0, b, a

Strategy


The following SCCs where found

b#(0, a(1, a(x, y))) → a#(0, a(x, y))a#(0, b(0, x)) → b#(0, a(0, x))
b#(0, a(1, a(x, y))) → a#(x, y)a#(0, a(1, a(x, y))) → a#(x, y)
a#(0, x) → b#(0, b(0, x))a#(0, a(x, y)) → a#(x, y)
a#(0, x) → b#(0, x)a#(0, a(1, a(x, y))) → a#(0, a(x, y))
a#(0, b(0, x)) → a#(0, x)

Problem 2: PolynomialLinearRange8NegiUR



Dependency Pair Problem

Dependency Pairs

b#(0, a(1, a(x, y)))a#(0, a(x, y))a#(0, b(0, x))b#(0, a(0, x))
b#(0, a(1, a(x, y)))a#(x, y)a#(0, a(1, a(x, y)))a#(x, y)
a#(0, x)b#(0, b(0, x))a#(0, a(x, y))a#(x, y)
a#(0, x)b#(0, x)a#(0, a(1, a(x, y)))a#(0, a(x, y))
a#(0, b(0, x))a#(0, x)

Rewrite Rules

a(0, b(0, x))b(0, a(0, x))a(0, x)b(0, b(0, x))
a(0, a(1, a(x, y)))a(1, a(0, a(x, y)))b(0, a(1, a(x, y)))b(1, a(0, a(x, y)))
a(0, a(x, y))a(1, a(1, a(x, y)))

Original Signature

Termination of terms over the following signature is verified: 1, 0, b, a

Strategy


Polynomial Interpretation

Improved Usable rules

a(0, a(1, a(x, y)))a(1, a(0, a(x, y)))a(0, b(0, x))b(0, a(0, x))
a(0, a(x, y))a(1, a(1, a(x, y)))a(0, x)b(0, b(0, x))
b(0, a(1, a(x, y)))b(1, a(0, a(x, y)))

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

a#(0, a(1, a(x, y)))a#(x, y)a#(0, b(0, x))a#(0, x)
a#(0, a(1, a(x, y)))a#(0, a(x, y))

Problem 3: PolynomialLinearRange8NegiUR



Dependency Pair Problem

Dependency Pairs

b#(0, a(1, a(x, y)))a#(0, a(x, y))a#(0, b(0, x))b#(0, a(0, x))
b#(0, a(1, a(x, y)))a#(x, y)a#(0, a(x, y))a#(x, y)
a#(0, x)b#(0, b(0, x))a#(0, x)b#(0, x)

Rewrite Rules

a(0, b(0, x))b(0, a(0, x))a(0, x)b(0, b(0, x))
a(0, a(1, a(x, y)))a(1, a(0, a(x, y)))b(0, a(1, a(x, y)))b(1, a(0, a(x, y)))
a(0, a(x, y))a(1, a(1, a(x, y)))

Original Signature

Termination of terms over the following signature is verified: 1, 0, b, a

Strategy


Polynomial Interpretation

Improved Usable rules

a(0, a(1, a(x, y)))a(1, a(0, a(x, y)))a(0, b(0, x))b(0, a(0, x))
a(0, a(x, y))a(1, a(1, a(x, y)))a(0, x)b(0, b(0, x))
b(0, a(1, a(x, y)))b(1, a(0, a(x, y)))

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

b#(0, a(1, a(x, y)))a#(x, y)a#(0, a(x, y))a#(x, y)

Problem 4: PolynomialLinearRange8NegiUR



Dependency Pair Problem

Dependency Pairs

b#(0, a(1, a(x, y)))a#(0, a(x, y))a#(0, b(0, x))b#(0, a(0, x))
a#(0, x)b#(0, b(0, x))a#(0, x)b#(0, x)

Rewrite Rules

a(0, b(0, x))b(0, a(0, x))a(0, x)b(0, b(0, x))
a(0, a(1, a(x, y)))a(1, a(0, a(x, y)))b(0, a(1, a(x, y)))b(1, a(0, a(x, y)))
a(0, a(x, y))a(1, a(1, a(x, y)))

Original Signature

Termination of terms over the following signature is verified: 1, 0, b, a

Strategy


Polynomial Interpretation

Improved Usable rules

a(0, a(1, a(x, y)))a(1, a(0, a(x, y)))a(0, b(0, x))b(0, a(0, x))
a(0, a(x, y))a(1, a(1, a(x, y)))a(0, x)b(0, b(0, x))
b(0, a(1, a(x, y)))b(1, a(0, a(x, y)))

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

a#(0, x)b#(0, x)

Problem 5: PolynomialLinearRange8NegiUR



Dependency Pair Problem

Dependency Pairs

b#(0, a(1, a(x, y)))a#(0, a(x, y))a#(0, b(0, x))b#(0, a(0, x))
a#(0, x)b#(0, b(0, x))

Rewrite Rules

a(0, b(0, x))b(0, a(0, x))a(0, x)b(0, b(0, x))
a(0, a(1, a(x, y)))a(1, a(0, a(x, y)))b(0, a(1, a(x, y)))b(1, a(0, a(x, y)))
a(0, a(x, y))a(1, a(1, a(x, y)))

Original Signature

Termination of terms over the following signature is verified: 1, 0, b, a

Strategy


Polynomial Interpretation

Improved Usable rules

a(0, a(1, a(x, y)))a(1, a(0, a(x, y)))a(0, b(0, x))b(0, a(0, x))
a(0, a(x, y))a(1, a(1, a(x, y)))a(0, x)b(0, b(0, x))
b(0, a(1, a(x, y)))b(1, a(0, a(x, y)))

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

a#(0, x)b#(0, b(0, x))