NO
The TRS could be proven non-terminating. The proof took 178 ms.
The following reduction sequence is a witness for non-termination:
c# →* c#Problem 1 was processed with processor DependencyGraph (5ms). | Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (7ms), PolynomialLinearRange4iUR (51ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (30ms), DependencyGraph (0ms), ReductionPairSAT (9ms), DependencyGraph (1ms), SizeChangePrinciple (1ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (1ms)].
c# | → | c# | c# | → | f#(g(c)) |
c | → | f(g(c)) | f(g(X)) | → | g(X) |
Termination of terms over the following signature is verified: f, g, c
c# → c# |