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 (45ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (5ms), PolynomialLinearRange4iUR (419ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (3599ms), DependencyGraph (3ms), ReductionPairSAT (3548ms), DependencyGraph (2ms), SizeChangePrinciple (31ms), ForwardNarrowing (0ms), BackwardInstantiation (2ms), ForwardInstantiation (1ms), Propagation (3ms)].
 | – Problem 3 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

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

Rewrite Rules

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

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

Rewrite Rules

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

gr#(s(x), s(y)) → gr#(x, y)

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

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

gr#(s(x), s(y)) → gr#(x, y)

Rewrite Rules

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