YES

We show the termination of the TRS R:

  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'|()))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(g(i(a(),b(),|b'|()),c()),d()) -> f#(.(b(),c()),|d'|())
p2: f#(g(i(a(),b(),|b'|()),c()),d()) -> f#(.(|b'|(),c()),|d'|())
p3: f#(g(h(a(),b()),c()),d()) -> f#(.(b(),g(h(a(),b()),c())),d())
p4: f#(g(h(a(),b()),c()),d()) -> f#(c(),|d'|())

and R consists of:

r1: f(g(i(a(),b(),|b'|()),c()),d()) -> if(e(),f(.(b(),c()),|d'|()),f(.(|b'|(),c()),|d'|()))
r2: f(g(h(a(),b()),c()),d()) -> if(e(),f(.(b(),g(h(a(),b()),c())),d()),f(c(),|d'|()))

The estimated dependency graph contains the following SCCs:

  (no SCCs)