TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60000 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (61ms).
 | – Problem 2 was processed with processor SubtermCriterion (2ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 | – Problem 5 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (417ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (5787ms), DependencyGraph (1ms), ReductionPairSAT (391ms), DependencyGraph (1ms), SizeChangePrinciple (168ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (0ms), Propagation (1ms)].

The following open problems remain:



Open Dependency Pair Problem 5

Dependency Pairs

f#(s(x))f#(-(*(s(s(0)), s(x)), s(s(x))))

Rewrite Rules

-(x, 0)x-(s(x), s(y))-(x, y)
+(0, y)y+(s(x), y)s(+(x, y))
*(x, 0)0*(x, s(y))+(x, *(x, y))
f(s(x))f(-(*(s(s(0)), s(x)), s(s(x))))

Original Signature

Termination of terms over the following signature is verified: f, 0, s, *, +, -


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

*#(x, s(y))+#(x, *(x, y))*#(x, s(y))*#(x, y)
f#(s(x))f#(-(*(s(s(0)), s(x)), s(s(x))))f#(s(x))*#(s(s(0)), s(x))
+#(s(x), y)+#(x, y)-#(s(x), s(y))-#(x, y)
f#(s(x))-#(*(s(s(0)), s(x)), s(s(x)))

Rewrite Rules

-(x, 0)x-(s(x), s(y))-(x, y)
+(0, y)y+(s(x), y)s(+(x, y))
*(x, 0)0*(x, s(y))+(x, *(x, y))
f(s(x))f(-(*(s(s(0)), s(x)), s(s(x))))

Original Signature

Termination of terms over the following signature is verified: f, 0, s, *, +, -

Strategy


The following SCCs where found

*#(x, s(y)) → *#(x, y)

f#(s(x)) → f#(-(*(s(s(0)), s(x)), s(s(x))))

+#(s(x), y) → +#(x, y)

-#(s(x), s(y)) → -#(x, y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

+#(s(x), y)+#(x, y)

Rewrite Rules

-(x, 0)x-(s(x), s(y))-(x, y)
+(0, y)y+(s(x), y)s(+(x, y))
*(x, 0)0*(x, s(y))+(x, *(x, y))
f(s(x))f(-(*(s(s(0)), s(x)), s(s(x))))

Original Signature

Termination of terms over the following signature is verified: f, 0, s, *, +, -

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

+#(s(x), y)+#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

-#(s(x), s(y))-#(x, y)

Rewrite Rules

-(x, 0)x-(s(x), s(y))-(x, y)
+(0, y)y+(s(x), y)s(+(x, y))
*(x, 0)0*(x, s(y))+(x, *(x, y))
f(s(x))f(-(*(s(s(0)), s(x)), s(s(x))))

Original Signature

Termination of terms over the following signature is verified: f, 0, s, *, +, -

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

-#(s(x), s(y))-#(x, y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

*#(x, s(y))*#(x, y)

Rewrite Rules

-(x, 0)x-(s(x), s(y))-(x, y)
+(0, y)y+(s(x), y)s(+(x, y))
*(x, 0)0*(x, s(y))+(x, *(x, y))
f(s(x))f(-(*(s(s(0)), s(x)), s(s(x))))

Original Signature

Termination of terms over the following signature is verified: f, 0, s, *, +, -

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

*#(x, s(y))*#(x, y)