YES
We show the termination of the TRS R:

  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  f(|0|()) -> s(|0|())
  f(s(x)) -> minus(s(x),g(f(x)))
  g(|0|()) -> |0|()
  g(s(x)) -> minus(s(x),f(g(x)))


-- SCC decomposition.

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

p1: minus#(s(x),s(y)) -> minus#(x,y)
p2: f#(s(x)) -> minus#(s(x),g(f(x)))
p3: f#(s(x)) -> g#(f(x))
p4: f#(s(x)) -> f#(x)
p5: g#(s(x)) -> minus#(s(x),f(g(x)))
p6: g#(s(x)) -> f#(g(x))
p7: g#(s(x)) -> g#(x)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: f(|0|()) -> s(|0|())
r4: f(s(x)) -> minus(s(x),g(f(x)))
r5: g(|0|()) -> |0|()
r6: g(s(x)) -> minus(s(x),f(g(x)))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p6, p7}
  {p1}


-- Reduction pair.

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

p1: g#(s(x)) -> g#(x)
p2: g#(s(x)) -> f#(g(x))
p3: f#(s(x)) -> f#(x)
p4: f#(s(x)) -> g#(f(x))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: f(|0|()) -> s(|0|())
r4: f(s(x)) -> minus(s(x),g(f(x)))
r5: g(|0|()) -> |0|()
r6: g(s(x)) -> minus(s(x),f(g(x)))

The set of usable rules consists of

  r1, r2, r3, r4, r5, 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              g#_A(x1) = x1
              s_A(x1) = x1 + 7
              f#_A(x1) = x1 + 4
              g_A(x1) = x1 + 1
              f_A(x1) = x1 + 11
              minus_A(x1,x2) = x1 + 1
              |0|_A() = 1
              g#!_A(x1) = x1
              s!_A(x1) = 10
              f#!_A(x1) = x1 + 5
              g!_A(x1) = 8
              f!_A(x1) = 11
              minus!_A(x1,x2) = 7
              |0|!_A() = 9
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g#_A(x1) = max{0, x1 - 23}
            s_A(x1) = max{22, x1 + 11}
            f#_A(x1) = max{0, x1 - 23}
            g_A(x1) = x1 + 2
            f_A(x1) = max{12, x1 + 11}
            minus_A(x1,x2) = max{23, x1 + 2, x2 - 15}
            |0|_A = 22
            g#!_A(x1) = max{5, x1 - 13}
            s!_A(x1) = 3
            f#!_A(x1) = max{4, x1 - 5}
            g!_A(x1) = 1
            f!_A(x1) = max{7, x1 - 23}
            minus!_A(x1,x2) = max{2, x1 - 33}
            |0|!_A = 2
        
  
    partial status:
  
      pi(g#) = [1]
      pi(s) = [1]
      pi(f#) = [1]
      pi(g) = [1]
      pi(f) = [1]
      pi(minus) = [1]
      pi(|0|) = []

The following rules are removed:

  p1, p2, p3, p4

Then no dependency pair remains.

-- Reduction pair.

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

p1: minus#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: f(|0|()) -> s(|0|())
r4: f(s(x)) -> minus(s(x),g(f(x)))
r5: g(|0|()) -> |0|()
r6: g(s(x)) -> minus(s(x),f(g(x)))

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              minus#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              minus#!_A(x1,x2) = x1 + x2 + 2
              s!_A(x1) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = 0
            s_A(x1) = max{8, x1 - 21}
            minus#!_A(x1,x2) = max{5, x1 - 37}
            s!_A(x1) = max{4, x1 - 36}
        
  
    partial status:
  
      pi(minus#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.