MAYBE

The TRS could not be proven terminating. The proof attempt took 29490 ms.

The following DP Processors were used


Problem 1 was processed with processor PolynomialLinearRange4iUR (0ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4iUR (0ms).
 |    | – Problem 3 was processed with processor PolynomialLinearRange4iUR (0ms).
 |    |    | – Problem 4 remains open; application of the following processors failed [DependencyGraph (2ms), PolynomialLinearRange4iUR (278ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (1834ms), DependencyGraph (2ms), ReductionPairSAT (25884ms), DependencyGraph (2ms), SizeChangePrinciple (8ms)].

The following open problems remain:



Open Dependency Pair Problem 4

Dependency Pairs

f#(f(f(a, b), c), x) → f#(a, f(c, f(b, x)))f#(x, f(y, z)) → f#(x, y)
f#(x, f(y, z)) → f#(f(x, y), z)

Rewrite Rules

f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))f(x, f(y, z)) → f(f(x, y), z)

Original Signature

Termination of terms over the following signature is verified: f, b, c, a


Problem 1: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

f#(f(f(a, b), c), x) → f#(a, f(c, f(b, x)))f#(x, f(y, z)) → f#(x, y)
f#(f(f(a, b), c), x) → f#(b, x)f#(f(f(a, b), c), x) → f#(c, f(b, x))
f#(x, f(y, z)) → f#(f(x, y), z)f#(f(f(a, b), c), x) → f#(b, f(a, f(c, f(b, x))))

Rewrite Rules

f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))f(x, f(y, z)) → f(f(x, y), z)

Original Signature

Termination of terms over the following signature is verified: f, b, c, a

Strategy


Polynomial Interpretation

Improved Usable rules

f(x, f(y, z)) → f(f(x, y), z)f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))

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

f#(f(f(a, b), c), x) → f#(b, x)

Problem 2: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

f#(f(f(a, b), c), x) → f#(a, f(c, f(b, x)))f#(x, f(y, z)) → f#(x, y)
f#(f(f(a, b), c), x) → f#(c, f(b, x))f#(x, f(y, z)) → f#(f(x, y), z)
f#(f(f(a, b), c), x) → f#(b, f(a, f(c, f(b, x))))

Rewrite Rules

f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))f(x, f(y, z)) → f(f(x, y), z)

Original Signature

Termination of terms over the following signature is verified: f, b, c, a

Strategy


Polynomial Interpretation

Improved Usable rules

f(x, f(y, z)) → f(f(x, y), z)f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))

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

f#(f(f(a, b), c), x) → f#(b, f(a, f(c, f(b, x))))

Problem 3: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

f#(f(f(a, b), c), x) → f#(a, f(c, f(b, x)))f#(x, f(y, z)) → f#(x, y)
f#(f(f(a, b), c), x) → f#(c, f(b, x))f#(x, f(y, z)) → f#(f(x, y), z)

Rewrite Rules

f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))f(x, f(y, z)) → f(f(x, y), z)

Original Signature

Termination of terms over the following signature is verified: f, b, c, a

Strategy


Polynomial Interpretation

Improved Usable rules

f(x, f(y, z)) → f(f(x, y), z)f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))

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

f#(f(f(a, b), c), x) → f#(c, f(b, x))