MAYBE

The TRS could not be proven terminating. The proof attempt took 15009 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (0ms).
 | – Problem 2 was processed with processor SubtermCriterion (0ms).
 | – Problem 3 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (9ms), PolynomialLinearRange4iUR (602ms), DependencyGraph (5ms), PolynomialLinearRange8NegiUR (2683ms), DependencyGraph (5ms), ReductionPairSAT (11368ms), DependencyGraph (5ms), SizeChangePrinciple (81ms)].

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

times#(x, plus(y, s(z)))times#(x, plus(y, times(s(z), 0)))times#(x, plus(y, s(z)))times#(x, s(z))
times#(x, plus(y, s(z)))times#(s(z), 0)times#(x, s(y))times#(x, y)

Rewrite Rules

times(x, plus(y, s(z)))plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))times(x, 0)0
times(x, s(y))plus(times(x, y), x)plus(x, 0)x
plus(x, s(y))s(plus(x, y))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, times


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

times#(x, plus(y, s(z)))times#(x, plus(y, times(s(z), 0)))times#(x, plus(y, s(z)))times#(x, s(z))
times#(x, plus(y, s(z)))plus#(y, times(s(z), 0))times#(x, s(y))plus#(times(x, y), x)
plus#(x, s(y))plus#(x, y)times#(x, plus(y, s(z)))times#(s(z), 0)
times#(x, s(y))times#(x, y)times#(x, plus(y, s(z)))plus#(times(x, plus(y, times(s(z), 0))), times(x, s(z)))

Rewrite Rules

times(x, plus(y, s(z)))plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))times(x, 0)0
times(x, s(y))plus(times(x, y), x)plus(x, 0)x
plus(x, s(y))s(plus(x, y))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, times

Strategy


The following SCCs where found

plus#(x, s(y)) → plus#(x, y)

times#(x, plus(y, s(z))) → times#(x, plus(y, times(s(z), 0)))times#(x, plus(y, s(z))) → times#(x, s(z))
times#(x, plus(y, s(z))) → times#(s(z), 0)times#(x, s(y)) → times#(x, y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

plus#(x, s(y))plus#(x, y)

Rewrite Rules

times(x, plus(y, s(z)))plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))times(x, 0)0
times(x, s(y))plus(times(x, y), x)plus(x, 0)x
plus(x, s(y))s(plus(x, y))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, times

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

plus#(x, s(y))plus#(x, y)