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) = max{82, x1 + 21}
            half_A(x1) = max{64, x1 + 13}
            double_A(x1) = max{63, x1 + 1}
            g_A(x1,x2) = max{64, x1 - 78, x2 + 2}
            |0|_A = 101
            h_A = 0
            f#!_A(x1,x2) = max{29, x1 - 65}
            s!_A(x1) = max{31, x1 - 29}
            half!_A(x1) = max{16, x1 - 66}
            double!_A(x1) = max{30, x1}
            g!_A(x1,x2) = max{1, x1 - 1, x2 - 67}
            |0|!_A = 33
            h!_A = 34
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1,x2) = x1
              s_A(x1) = x1 + 3
              half_A(x1) = x1
              double_A(x1) = 1
              g_A(x1,x2) = x2
              |0|_A() = 1
              h_A() = 1
              f#!_A(x1,x2) = x1 + 1
              s!_A(x1) = x1 + 5
              half!_A(x1) = x1
              double!_A(x1) = 3
              g!_A(x1,x2) = x2 + 7
              |0|!_A() = 5
              h!_A() = 6
        
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      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

  (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{16, x1 - 28}
            g#!_A(x1,x2) = 4
            d!_A = 8
            s!_A(x1) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              g#_A(x1,x2) = x2
              d_A() = 1
              s_A(x1) = x1 + 2
              g#!_A(x1,x2) = x2
              d!_A() = 8
              s!_A(x1) = x1 + 1
        
  
    partial status:
  
      pi(g#) = []
      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

  (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
            h_A = 0
            s_A(x1) = max{8, x1 - 11}
            g#!_A(x1,x2) = max{1, x1 - 1}
            h!_A = 4
            s!_A(x1) = 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              g#_A(x1,x2) = x2
              h_A() = 1
              s_A(x1) = x1 + 3
              g#!_A(x1,x2) = x2 + 3
              h!_A() = 10
              s!_A(x1) = 2
        
  
    partial status:
  
      pi(g#) = []
      pi(h) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.