TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60119 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (47ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (5ms), PolynomialLinearRange4iUR (1174ms), DependencyGraph (4ms), PolynomialLinearRange8NegiUR (17733ms), DependencyGraph (2ms), ReductionPairSAT (1155ms), DependencyGraph (1ms), SizeChangePrinciple (171ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (0ms)].
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 | – Problem 5 was processed with processor SubtermCriterion (0ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

f#(s(x), s(y))f#(-(y, min(x, y)), s(twice(min(x, y))))f#(s(x), s(y))f#(-(x, min(x, y)), s(twice(min(x, y))))

Rewrite Rules

-(x, 0)x-(s(x), s(y))-(x, y)
min(x, 0)0min(0, y)0
min(s(x), s(y))s(min(x, y))twice(0)0
twice(s(x))s(s(twice(x)))f(s(x), s(y))f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y))f(-(x, min(x, y)), s(twice(min(x, y))))

Original Signature

Termination of terms over the following signature is verified: f, min, twice, 0, s, -


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

f#(s(x), s(y))f#(-(y, min(x, y)), s(twice(min(x, y))))f#(s(x), s(y))f#(-(x, min(x, y)), s(twice(min(x, y))))
f#(s(x), s(y))-#(x, min(x, y))f#(s(x), s(y))twice#(min(x, y))
min#(s(x), s(y))min#(x, y)f#(s(x), s(y))-#(y, min(x, y))
f#(s(x), s(y))min#(x, y)twice#(s(x))twice#(x)
-#(s(x), s(y))-#(x, y)

Rewrite Rules

-(x, 0)x-(s(x), s(y))-(x, y)
min(x, 0)0min(0, y)0
min(s(x), s(y))s(min(x, y))twice(0)0
twice(s(x))s(s(twice(x)))f(s(x), s(y))f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y))f(-(x, min(x, y)), s(twice(min(x, y))))

Original Signature

Termination of terms over the following signature is verified: min, f, 0, twice, s, -

Strategy


The following SCCs where found

min#(s(x), s(y)) → min#(x, y)

twice#(s(x)) → twice#(x)

-#(s(x), s(y)) → -#(x, y)

f#(s(x), s(y)) → f#(-(y, min(x, y)), s(twice(min(x, y))))f#(s(x), s(y)) → f#(-(x, min(x, y)), s(twice(min(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)
min(x, 0)0min(0, y)0
min(s(x), s(y))s(min(x, y))twice(0)0
twice(s(x))s(s(twice(x)))f(s(x), s(y))f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y))f(-(x, min(x, y)), s(twice(min(x, y))))

Original Signature

Termination of terms over the following signature is verified: min, f, 0, twice, 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

twice#(s(x))twice#(x)

Rewrite Rules

-(x, 0)x-(s(x), s(y))-(x, y)
min(x, 0)0min(0, y)0
min(s(x), s(y))s(min(x, y))twice(0)0
twice(s(x))s(s(twice(x)))f(s(x), s(y))f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y))f(-(x, min(x, y)), s(twice(min(x, y))))

Original Signature

Termination of terms over the following signature is verified: min, f, 0, twice, s, -

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

twice#(s(x))twice#(x)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

min#(s(x), s(y))min#(x, y)

Rewrite Rules

-(x, 0)x-(s(x), s(y))-(x, y)
min(x, 0)0min(0, y)0
min(s(x), s(y))s(min(x, y))twice(0)0
twice(s(x))s(s(twice(x)))f(s(x), s(y))f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y))f(-(x, min(x, y)), s(twice(min(x, y))))

Original Signature

Termination of terms over the following signature is verified: min, f, 0, twice, s, -

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

min#(s(x), s(y))min#(x, y)