YES

Problem:
 f(g(i(a(),b(),b'()),c()),d()) -> if(e(),f(.(b(),c()),d'()),f(.(b'(),c()),d'()))
 f(g(h(a(),b()),c()),d()) -> if(e(),f(.(b(),g(h(a(),b()),c())),d()),f(c(),d'()))

Proof:
 DP Processor:
  DPs:
   f#(g(i(a(),b(),b'()),c()),d()) -> f#(.(b'(),c()),d'())
   f#(g(i(a(),b(),b'()),c()),d()) -> f#(.(b(),c()),d'())
   f#(g(h(a(),b()),c()),d()) -> f#(c(),d'())
   f#(g(h(a(),b()),c()),d()) -> f#(.(b(),g(h(a(),b()),c())),d())
  TRS:
   f(g(i(a(),b(),b'()),c()),d()) -> if(e(),f(.(b(),c()),d'()),f(.(b'(),c()),d'()))
   f(g(h(a(),b()),c()),d()) -> if(e(),f(.(b(),g(h(a(),b()),c())),d()),f(c(),d'()))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0, x1) = 4x0 + 3x1 + 0,
    
    [h](x0, x1) = x1,
    
    [if](x0, x1, x2) = x2,
    
    [d'] = 1,
    
    [.](x0, x1) = x0 + 0,
    
    [e] = 2,
    
    [f](x0, x1) = 1x0 + x1 + 0,
    
    [d] = 0,
    
    [g](x0, x1) = x0 + 1x1 + 0,
    
    [c] = 0,
    
    [i](x0, x1, x2) = 2x2,
    
    [b'] = 0,
    
    [b] = 0,
    
    [a] = 2
   orientation:
    f#(g(i(a(),b(),b'()),c()),d()) = 6 >= 4 = f#(.(b'(),c()),d'())
    
    f#(g(i(a(),b(),b'()),c()),d()) = 6 >= 4 = f#(.(b(),c()),d'())
    
    f#(g(h(a(),b()),c()),d()) = 5 >= 4 = f#(c(),d'())
    
    f#(g(h(a(),b()),c()),d()) = 5 >= 4 = f#(.(b(),g(h(a(),b()),c())),d())
    
    f(g(i(a(),b(),b'()),c()),d()) = 3 >= 1 = if(e(),f(.(b(),c()),d'()),f(.(b'(),c()),d'()))
    
    f(g(h(a(),b()),c()),d()) = 2 >= 1 = if(e(),f(.(b(),g(h(a(),b()),c())),d()),f(c(),d'()))
   problem:
    DPs:
     
    TRS:
     f(g(i(a(),b(),b'()),c()),d()) -> if(e(),f(.(b(),c()),d'()),f(.(b'(),c()),d'()))
     f(g(h(a(),b()),c()),d()) -> if(e(),f(.(b(),g(h(a(),b()),c())),d()),f(c(),d'()))
   Qed