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 (48ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (5ms), PolynomialLinearRange4iUR (343ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (5346ms), DependencyGraph (4ms), ReductionPairSAT (1939ms), DependencyGraph (5ms), SizeChangePrinciple (127ms), ForwardNarrowing (1ms), BackwardInstantiation (2ms), ForwardInstantiation (1ms), Propagation (3ms)].
 | – Problem 4 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

cond2#(false, x, y) → cond1#(neq(x, 0), p(x), y)cond1#(true, x, y) → cond2#(gr(x, y), x, y)
cond2#(true, x, y) → cond1#(neq(x, 0), y, y)

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, y), x, y)cond2(true, x, y) → cond1(neq(x, 0), y, y)
cond2(false, x, y) → cond1(neq(x, 0), p(x), y)gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
neq(0, 0) → falseneq(0, s(x)) → true
neq(s(x), 0) → trueneq(s(x), s(y)) → neq(x, y)
p(0) → 0p(s(x)) → x

Original Signature

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


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, y), x, y)cond2(true, x, y) → cond1(neq(x, 0), y, y)
cond2(false, x, y) → cond1(neq(x, 0), p(x), y)gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
neq(0, 0) → falseneq(0, s(x)) → true
neq(s(x), 0) → trueneq(s(x), s(y)) → neq(x, y)
p(0) → 0p(s(x)) → x

Original Signature

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

Strategy


The following SCCs where found

neq#(s(x), s(y)) → neq#(x, y)

gr#(s(x), s(y)) → gr#(x, y)

cond2#(false, x, y) → cond1#(neq(x, 0), p(x), y)cond1#(true, x, y) → cond2#(gr(x, y), x, y)
cond2#(true, x, y) → cond1#(neq(x, 0), y, y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

neq#(s(x), s(y)) → neq#(x, y)

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, y), x, y)cond2(true, x, y) → cond1(neq(x, 0), y, y)
cond2(false, x, y) → cond1(neq(x, 0), p(x), y)gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
neq(0, 0) → falseneq(0, s(x)) → true
neq(s(x), 0) → trueneq(s(x), s(y)) → neq(x, y)
p(0) → 0p(s(x)) → x

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

neq#(s(x), s(y)) → neq#(x, y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

gr#(s(x), s(y)) → gr#(x, y)

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, y), x, y)cond2(true, x, y) → cond1(neq(x, 0), y, y)
cond2(false, x, y) → cond1(neq(x, 0), p(x), y)gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
neq(0, 0) → falseneq(0, s(x)) → true
neq(s(x), 0) → trueneq(s(x), s(y)) → neq(x, y)
p(0) → 0p(s(x)) → x

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

gr#(s(x), s(y)) → gr#(x, y)