TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60093 ms.

The following DP Processors were used


Problem 1 was processed with processor PolynomialLinearRange4iUR (7845ms).
 | – Problem 2 was processed with processor DependencyGraph (20ms).
 |    | – Problem 3 remains open; application of the following processors failed [PolynomialLinearRange4iUR (842ms), DependencyGraph (0ms), PolynomialLinearRange4iUR (797ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (851ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (9988ms), DependencyGraph (0ms)].
 |    | – Problem 4 was processed with processor PolynomialLinearRange4iUR (3505ms).
 |    |    | – Problem 5 was processed with processor PolynomialLinearRange4iUR (463ms).

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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


Problem 1: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy


Polynomial Interpretation

Improved Usable rules

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

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

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

Problem 2: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy


The following SCCs where found

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

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

Problem 4: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy


Polynomial Interpretation

Improved Usable rules

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

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

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

Problem 5: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Original Signature

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

Strategy


Polynomial Interpretation

Improved Usable rules

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

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

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