YES

Problem:
 +(a(),b()) -> +(b(),a())
 +(a(),+(b(),z)) -> +(b(),+(a(),z))
 +(+(x,y),z) -> +(x,+(y,z))
 f(a(),y) -> a()
 f(b(),y) -> b()
 f(+(x,y),z) -> +(f(x,z),f(y,z))

Proof:
 LPO Processor:
  precedence:
   f > a > + ~ b
  problem:
   
  Qed