TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60016 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (53ms).
 | – Problem 2 was processed with processor SubtermCriterion (2ms).
 | – Problem 3 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (5ms), PolynomialLinearRange4iUR (363ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (2644ms), DependencyGraph (4ms), ReductionPairSAT (2332ms), DependencyGraph (4ms), SizeChangePrinciple (72ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (3ms), Propagation (1ms)].

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

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

Rewrite Rules

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


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

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

Strategy


The following SCCs where found

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

gr#(s(x), s(y)) → gr#(x, y)

Problem 2: 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(gr(x, 0), y, y)
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(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, gr, cond1

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

gr#(s(x), s(y)) → gr#(x, y)