YES

Problem:
 h(f(f(c())),b()) -> f(h(h(h(c(),h(f(h(c(),f(b()))),a())),b()),c()))
 c() -> c()
 f(f(h(h(f(a()),a()),c()))) -> f(h(f(c()),b()))
 h(f(h(f(b()),h(h(f(h(c(),f(c()))),b()),a()))),h(a(),c())) -> c()

Proof:
 Church Rosser Transformation Processor (no redundant rules):
  strict:
   h(f(h(f(b()),h(h(f(h(c(),f(c()))),b()),a()))),h(a(),c())) -> c()
   f(f(h(h(f(a()),a()),c()))) -> f(h(f(c()),b()))
   h(f(f(c())),b()) -> f(h(h(h(c(),h(f(h(c(),f(b()))),a())),b()),c()))
  weak:
   
  critical peaks: 0
  Closedness Processor (*feeble*):
   
   Qed