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 (50ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (405ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (271ms), DependencyGraph (0ms), PolynomialLinearRange8NegiUR (3273ms), DependencyGraph (0ms), ReductionPairSAT (8676ms), DependencyGraph (1ms), SizeChangePrinciple (143ms), ForwardNarrowing (0ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (0ms)].
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 5 was processed with processor PolynomialLinearRange4iUR (102ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

h#(e(x), y) → h#(d(x, y), s(y))

Rewrite Rules

h(e(x), y) → h(d(x, y), s(y))d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))g(e(x), e(y)) → e(g(x, y))

Original Signature

Termination of terms over the following signature is verified: g, d, 0, e, s, h


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

h#(e(x), y) → h#(d(x, y), s(y))h#(e(x), y) → d#(x, y)
d#(g(g(0, x), y), s(z)) → g#(e(x), d(g(g(0, x), y), z))d#(g(x, y), z) → g#(d(x, z), e(y))
d#(g(x, y), z) → d#(x, z)d#(g(g(0, x), y), s(z)) → d#(g(g(0, x), y), z)
g#(e(x), e(y)) → g#(x, y)d#(g(g(0, x), y), s(z)) → g#(g(0, x), y)
d#(g(g(0, x), y), s(z)) → g#(0, x)

Rewrite Rules

h(e(x), y) → h(d(x, y), s(y))d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))g(e(x), e(y)) → e(g(x, y))

Original Signature

Termination of terms over the following signature is verified: g, d, e, 0, s, h

Strategy


The following SCCs where found

h#(e(x), y) → h#(d(x, y), s(y))

d#(g(x, y), z) → d#(x, z)d#(g(g(0, x), y), s(z)) → d#(g(g(0, x), y), z)

g#(e(x), e(y)) → g#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

d#(g(x, y), z) → d#(x, z)d#(g(g(0, x), y), s(z)) → d#(g(g(0, x), y), z)

Rewrite Rules

h(e(x), y) → h(d(x, y), s(y))d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))g(e(x), e(y)) → e(g(x, y))

Original Signature

Termination of terms over the following signature is verified: g, d, e, 0, s, h

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

d#(g(x, y), z) → d#(x, z)

Problem 5: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

d#(g(g(0, x), y), s(z)) → d#(g(g(0, x), y), z)

Rewrite Rules

h(e(x), y) → h(d(x, y), s(y))d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))g(e(x), e(y)) → e(g(x, y))

Original Signature

Termination of terms over the following signature is verified: g, d, 0, e, s, h

Strategy


Polynomial Interpretation

Improved Usable rules

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

d#(g(g(0, x), y), s(z)) → d#(g(g(0, x), y), z)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

g#(e(x), e(y)) → g#(x, y)

Rewrite Rules

h(e(x), y) → h(d(x, y), s(y))d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))g(e(x), e(y)) → e(g(x, y))

Original Signature

Termination of terms over the following signature is verified: g, d, e, 0, s, h

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

g#(e(x), e(y)) → g#(x, y)