YES
We show the termination of the TRS R:

  f(f(x)) -> f(c(f(x)))
  f(f(x)) -> f(d(f(x)))
  g(c(x)) -> x
  g(d(x)) -> x
  g(c(h(|0|()))) -> g(d(|1|()))
  g(c(|1|())) -> g(d(h(|0|())))
  g(h(x)) -> g(x)


-- SCC decomposition.

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

p1: f#(f(x)) -> f#(c(f(x)))
p2: f#(f(x)) -> f#(d(f(x)))
p3: g#(c(h(|0|()))) -> g#(d(|1|()))
p4: g#(c(|1|())) -> g#(d(h(|0|())))
p5: g#(h(x)) -> g#(x)

and R consists of:

r1: f(f(x)) -> f(c(f(x)))
r2: f(f(x)) -> f(d(f(x)))
r3: g(c(x)) -> x
r4: g(d(x)) -> x
r5: g(c(h(|0|()))) -> g(d(|1|()))
r6: g(c(|1|())) -> g(d(h(|0|())))
r7: g(h(x)) -> g(x)

The estimated dependency graph contains the following SCCs:

  {p5}


-- Reduction pair.

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

p1: g#(h(x)) -> g#(x)

and R consists of:

r1: f(f(x)) -> f(c(f(x)))
r2: f(f(x)) -> f(d(f(x)))
r3: g(c(x)) -> x
r4: g(d(x)) -> x
r5: g(c(h(|0|()))) -> g(d(|1|()))
r6: g(c(|1|())) -> g(d(h(|0|())))
r7: g(h(x)) -> g(x)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            g#_A(0) = 0
            g#_A(1) = 0
            h_A(0) = 0
            h_A(1) = 1
            f_A(0) = 0
            f_A(1) = 0
            c_A(0) = 1
            c_A(1) = 0
            d_A(0) = 1
            d_A(1) = 0
            g_A(0) = 1
            g_A(1) = 0
            |0|_A = 0
            |1|_A = 0
            g#!_A(0) = 0
            g#!_A(1) = 0
            h!_A(0) = 0
            h!_A(1) = 0
            f!_A(0) = 0
            f!_A(1) = 0
            c!_A(0) = 1
            c!_A(1) = 0
            d!_A(0) = 0
            d!_A(1) = 0
            g!_A(0) = 0
            g!_A(1) = 0
            |0|!_A = 0
            |1|!_A = 1
  
    partial status:
  
      pi(g#) = [1]
      pi(h) = [1]
      pi(f) = []
      pi(c) = [1]
      pi(d) = [1]
      pi(g) = [1]
      pi(|0|) = []
      pi(|1|) = []

The following rules are removed:

  p1

Then no dependency pair remains.