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:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1,x2) = x1 + 10
        s_A(x1) = max{4, x1 - 3}
        half_A(x1) = 4
        double_A(x1) = max{6, x1 + 5}
        g_A(x1,x2) = max{2, x1 - 2}
        |0|_A = 2
        h_A = 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1,x2) = max{1, x1 - 7}
          s_A(x1) = x1 + 11
          half_A(x1) = max{8, x1 - 2}
          double_A(x1) = max{7, x1 + 2}
          g_A(x1,x2) = max{8, x1 - 24, x2 - 2}
          |0|_A = 0
          h_A = 23
          f#!_A(x1,x2) = max{10, x1 + 1}
          s!_A(x1) = max{3, x1 - 14}
          half!_A(x1) = max{11, x1 + 2}
          double!_A(x1) = max{13, x1 + 5}
          g!_A(x1,x2) = max{4, x1 - 25, x2 - 1}
          |0|!_A = 0
          h!_A = 22
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = []
      pi(half) = []
      pi(double) = [1]
      pi(g) = []
      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

  r1, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        g#_A(x1,x2) = max{6, x1 - 31, x2 + 2}
        d_A = 30
        s_A(x1) = max{27, x1 + 5}
        g_A(x1,x2) = max{31, x1 - 42, x2 - 33}
        |0|_A = 0
        h_A = 11
  
    base order 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{4, x1 - 14}
          g_A(x1,x2) = max{13, x1 - 16, x2 - 7}
          |0|_A = 6
          h_A = 14
          g#!_A(x1,x2) = x1 + 1
          d!_A = 1
          s!_A(x1) = max{4, x1 - 14}
          g!_A(x1,x2) = max{3, x1 - 45, x2 - 47}
          |0|!_A = 0
          h!_A = 1
  
    partial status:
  
      pi(g#) = [2]
      pi(d) = []
      pi(s) = []
      pi(g) = []
      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#(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:
  
      max/plus interpretation on natural numbers:
      
        g#_A(x1,x2) = max{x1 - 1, x2 + 11}
        h_A = 12
        s_A(x1) = x1 + 12
        g_A(x1,x2) = max{12, x1 - 51, x2 - 25}
        |0|_A = 11
  
    base order 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{6, x1 - 18}
          g_A(x1,x2) = max{16, x1 - 21, x2 - 12}
          |0|_A = 8
          g#!_A(x1,x2) = 0
          h!_A = 0
          s!_A(x1) = max{4, x1 - 20}
          g!_A(x1,x2) = max{2, x1 - 21, x2 - 21}
          |0|!_A = 0
  
    partial status:
  
      pi(g#) = [2]
      pi(h) = []
      pi(s) = []
      pi(g) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.