YES
We show the termination of the TRS R:

  g(x,|0|()) -> |0|()
  g(d(),s(x)) -> s(s(g(d(),x)))
  g(h(),s(|0|())) -> |0|()
  g(h(),s(s(x))) -> s(g(h(),x))
  double(x) -> g(d(),x)
  half(x) -> g(h(),x)
  f(s(x),y) -> f(half(s(x)),double(y))
  f(s(|0|()),y) -> y
  id(x) -> f(x,s(|0|()))


-- SCC decomposition.

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

p1: g#(d(),s(x)) -> g#(d(),x)
p2: g#(h(),s(s(x))) -> g#(h(),x)
p3: double#(x) -> g#(d(),x)
p4: half#(x) -> g#(h(),x)
p5: f#(s(x),y) -> f#(half(s(x)),double(y))
p6: f#(s(x),y) -> half#(s(x))
p7: f#(s(x),y) -> double#(y)
p8: id#(x) -> f#(x,s(|0|()))

and R consists of:

r1: g(x,|0|()) -> |0|()
r2: g(d(),s(x)) -> s(s(g(d(),x)))
r3: g(h(),s(|0|())) -> |0|()
r4: g(h(),s(s(x))) -> s(g(h(),x))
r5: double(x) -> g(d(),x)
r6: half(x) -> g(h(),x)
r7: f(s(x),y) -> f(half(s(x)),double(y))
r8: f(s(|0|()),y) -> y
r9: id(x) -> f(x,s(|0|()))

The estimated dependency graph contains the following SCCs:

  {p5}
  {p1}
  {p2}


-- Reduction pair.

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

p1: f#(s(x),y) -> f#(half(s(x)),double(y))

and R consists of:

r1: g(x,|0|()) -> |0|()
r2: g(d(),s(x)) -> s(s(g(d(),x)))
r3: g(h(),s(|0|())) -> |0|()
r4: g(h(),s(s(x))) -> s(g(h(),x))
r5: double(x) -> g(d(),x)
r6: half(x) -> g(h(),x)
r7: f(s(x),y) -> f(half(s(x)),double(y))
r8: f(s(|0|()),y) -> y
r9: id(x) -> f(x,s(|0|()))

The set of usable rules consists of

  r1, r3, r4, r6

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:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1,x2) = 0
            s_A(x1) = 5
            half_A(x1) = max{5, x1}
            double_A(x1) = x1 + 5
            g_A(x1,x2) = max{5, x1 - 12}
            |0|_A = 5
            h_A = 15
            f#!_A(x1,x2) = max{7, x1 + 3}
            s!_A(x1) = x1 + 12
            half!_A(x1) = max{14, x1 - 7}
            double!_A(x1) = 23
            g!_A(x1,x2) = max{10, x1 - 1}
            |0|!_A = 7
            h!_A = 18
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1,x2) = 0
            s_A(x1) = max{8, x1 + 5}
            half_A(x1) = max{4, x1 - 3}
            double_A(x1) = 10
            g_A(x1,x2) = max{4, x2 - 4}
            |0|_A = 3
            h_A = 9
            f#!_A(x1,x2) = max{10, x1 + 3}
            s!_A(x1) = 5
            half!_A(x1) = max{4, x1 + 1}
            double!_A(x1) = max{12, x1 + 1}
            g!_A(x1,x2) = max{2, x1 - 6, x2 - 13}
            |0|!_A = 7
            h!_A = 6
        
  
    partial status:
  
      pi(f#) = []
      pi(s) = []
      pi(half) = [1]
      pi(double) = [1]
      pi(g) = [2]
      pi(|0|) = []
      pi(h) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: g#(d(),s(x)) -> g#(d(),x)

and R consists of:

r1: g(x,|0|()) -> |0|()
r2: g(d(),s(x)) -> s(s(g(d(),x)))
r3: g(h(),s(|0|())) -> |0|()
r4: g(h(),s(s(x))) -> s(g(h(),x))
r5: double(x) -> g(d(),x)
r6: half(x) -> g(h(),x)
r7: f(s(x),y) -> f(half(s(x)),double(y))
r8: f(s(|0|()),y) -> y
r9: id(x) -> f(x,s(|0|()))

The set of usable rules consists of

  (no rules)

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:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g#_A(x1,x2) = 0
            d_A = 0
            s_A(x1) = max{3, x1 - 11}
            g#!_A(x1,x2) = max{2, x1 - 22, x2 - 24}
            d!_A = 1
            s!_A(x1) = max{4, x1 - 23}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g#_A(x1,x2) = 0
            d_A = 0
            s_A(x1) = max{7, x1 - 14}
            g#!_A(x1,x2) = max{2, x1 - 58, x2 - 43}
            d!_A = 1
            s!_A(x1) = max{4, x1 - 42}
        
  
    partial status:
  
      pi(g#) = [1, 2]
      pi(d) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

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

and R consists of:

r1: g(x,|0|()) -> |0|()
r2: g(d(),s(x)) -> s(s(g(d(),x)))
r3: g(h(),s(|0|())) -> |0|()
r4: g(h(),s(s(x))) -> s(g(h(),x))
r5: double(x) -> g(d(),x)
r6: half(x) -> g(h(),x)
r7: f(s(x),y) -> f(half(s(x)),double(y))
r8: f(s(|0|()),y) -> y
r9: id(x) -> f(x,s(|0|()))

The set of usable rules consists of

  r1, r3

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:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g#_A(x1,x2) = 0
            h_A = 68
            s_A(x1) = max{4, x1 - 12}
            g_A(x1,x2) = max{11, x1 - 70, x2 - 7}
            |0|_A = 1
            g#!_A(x1,x2) = max{23, x1 - 132, x2 - 39}
            h!_A = 24
            s!_A(x1) = max{29, x1 - 26}
            g!_A(x1,x2) = max{26, x1 - 39, x2 + 25}
            |0|!_A = 27
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g#_A(x1,x2) = 0
            h_A = 16
            s_A(x1) = max{3, x1 - 15}
            g_A(x1,x2) = max{14, x1 - 18, x2 - 6}
            |0|_A = 5
            g#!_A(x1,x2) = max{1, x1 - 17, x2 - 50}
            h!_A = 3
            s!_A(x1) = max{2, x1 - 34}
            g!_A(x1,x2) = max{4, x1 - 27, x2 - 11}
            |0|!_A = 0
        
  
    partial status:
  
      pi(g#) = [1, 2]
      pi(h) = []
      pi(s) = [1]
      pi(g) = [2]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.