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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1,x2) = max{27, x1}
        s_A(x1) = x1 + 27
        half_A(x1) = max{26, x1 - 11}
        double_A(x1) = max{13, x1 - 56}
        g_A(x1,x2) = max{14, x1 - 56, x2 - 12}
        |0|_A = 5
        h_A = 6
  
    base order 2:
  
      precedence:
      
        half > s = double = g > f# = h > |0|
      
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(half) = []
      pi(double) = []
      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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        g#_A(x1,x2) = max{6, x2 + 4}
        d_A = 14
        s_A(x1) = max{17, x1 + 5}
        g_A(x1,x2) = 11
        |0|_A = 8
        h_A = 23
  
    base order 2:
  
      precedence:
      
        h > s > g# > d > g > |0|
      
  
    partial status:
  
      pi(g#) = [2]
      pi(d) = []
      pi(s) = [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#(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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        g#_A(x1,x2) = max{2, x2 - 1}
        h_A = 17
        s_A(x1) = x1 + 12
        g_A(x1,x2) = max{12, x1 - 37, x2 - 23}
        |0|_A = 8
  
    base order 2:
  
      precedence:
      
        h > g# = s > g > |0|
      
  
    partial status:
  
      pi(g#) = []
      pi(h) = []
      pi(s) = [1]
      pi(g) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.