TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60001 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (80ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (5ms), PolynomialLinearRange4iUR (514ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (4991ms), DependencyGraph (45ms), ReductionPairSAT (541ms), DependencyGraph (3ms), SizeChangePrinciple (152ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (2ms), Propagation (1ms)].

The following open problems remain:



Open Dependency Pair Problem 4

Dependency Pairs

cond2#(true, x) → cond1#(neq(x, 0), div2(x))cond1#(true, x) → cond2#(even(x), x)
cond2#(false, x) → cond1#(neq(x, 0), p(x))

Rewrite Rules

cond1(true, x) → cond2(even(x), x)cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))neq(0, 0) → false
neq(0, s(x)) → trueneq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)even(0) → true
even(s(0)) → falseeven(s(s(x))) → even(x)
div2(0) → 0div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))p(0) → 0
p(s(x)) → x

Original Signature

Termination of terms over the following signature is verified: cond2, 0, s, p, false, true, even, neq, y, div2, cond1


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

cond2#(true, x) → div2#(x)cond2#(true, x) → cond1#(neq(x, 0), div2(x))
cond1#(true, x) → cond2#(even(x), x)cond2#(true, x) → neq#(x, 0)
even#(s(s(x))) → even#(x)cond2#(false, x) → neq#(x, 0)
cond2#(false, x) → p#(x)cond1#(true, x) → even#(x)
cond2#(false, x) → cond1#(neq(x, 0), p(x))neq#(s(x), s(y)) → neq#(x, y)
div2#(s(s(x))) → div2#(x)

Rewrite Rules

cond1(true, x) → cond2(even(x), x)cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))neq(0, 0) → false
neq(0, s(x)) → trueneq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)even(0) → true
even(s(0)) → falseeven(s(s(x))) → even(x)
div2(0) → 0div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))p(0) → 0
p(s(x)) → x

Original Signature

Termination of terms over the following signature is verified: cond2, 0, s, p, true, false, even, neq, cond1, div2, y

Strategy


The following SCCs where found

even#(s(s(x))) → even#(x)

cond2#(true, x) → cond1#(neq(x, 0), div2(x))cond1#(true, x) → cond2#(even(x), x)
cond2#(false, x) → cond1#(neq(x, 0), p(x))

div2#(s(s(x))) → div2#(x)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

even#(s(s(x))) → even#(x)

Rewrite Rules

cond1(true, x) → cond2(even(x), x)cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))neq(0, 0) → false
neq(0, s(x)) → trueneq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)even(0) → true
even(s(0)) → falseeven(s(s(x))) → even(x)
div2(0) → 0div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))p(0) → 0
p(s(x)) → x

Original Signature

Termination of terms over the following signature is verified: cond2, 0, s, p, true, false, even, neq, cond1, div2, y

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

even#(s(s(x))) → even#(x)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

div2#(s(s(x))) → div2#(x)

Rewrite Rules

cond1(true, x) → cond2(even(x), x)cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))neq(0, 0) → false
neq(0, s(x)) → trueneq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)even(0) → true
even(s(0)) → falseeven(s(s(x))) → even(x)
div2(0) → 0div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))p(0) → 0
p(s(x)) → x

Original Signature

Termination of terms over the following signature is verified: cond2, 0, s, p, true, false, even, neq, cond1, div2, y

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

div2#(s(s(x))) → div2#(x)