MAYBE

Problem:
 a(b(x)) -> a(c(b(x)))

Proof:
 DP Processor:
  DPs:
   a#(b(x)) -> a#(c(b(x)))
  TRS:
   a(b(x)) -> a(c(b(x)))
  SCC Processor:
   #sccs: 1
   #rules: 1
   #arcs: 1/1
   DPs:
    a#(b(x)) -> a#(c(b(x)))
   TRS:
    a(b(x)) -> a(c(b(x)))
   Open