MAYBE
The TRS could not be proven terminating. The proof attempt took 11465 ms.
The following DP Processors were used
Problem 1 was processed with processor PolynomialLinearRange4iUR (0ms).
| Problem 2 remains open; application of the following processors failed [DependencyGraph (1ms), PolynomialLinearRange4iUR (196ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (1592ms), DependencyGraph (1ms), ReductionPairSAT (9040ms), DependencyGraph (0ms), SizeChangePrinciple (51ms)].
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
p#(a(x0), p(x1, p(x2, x3))) | → | p#(x1, p(x0, p(a(x3), x3))) |
Rewrite Rules
p(a(x0), p(x1, p(x2, x3))) | → | p(x1, p(x0, p(a(x3), x3))) |
Original Signature
Termination of terms over the following signature is verified: p, a
Problem 1: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
p#(a(x0), p(x1, p(x2, x3))) | → | p#(a(x3), x3) | | p#(a(x0), p(x1, p(x2, x3))) | → | p#(x0, p(a(x3), x3)) |
p#(a(x0), p(x1, p(x2, x3))) | → | p#(x1, p(x0, p(a(x3), x3))) |
Rewrite Rules
p(a(x0), p(x1, p(x2, x3))) | → | p(x1, p(x0, p(a(x3), x3))) |
Original Signature
Termination of terms over the following signature is verified: p, a
Strategy
Polynomial Interpretation
- a(x): 0
- p(x,y): y + 1
- p#(x,y): y
Improved Usable rules
p(a(x0), p(x1, p(x2, x3))) | → | p(x1, p(x0, p(a(x3), x3))) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
p#(a(x0), p(x1, p(x2, x3))) | → | p#(a(x3), x3) | | p#(a(x0), p(x1, p(x2, x3))) | → | p#(x0, p(a(x3), x3)) |