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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          g#_A(x1) = x1 + 1
          s_A(x1) = x1 + 5
          f#_A(x1) = x1 + 2
          g_A(x1) = x1 + 2
          f_A(x1) = x1 + 5
          minus_A(x1,x2) = x1 + 1
          |0|_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          g#_A(x1) = max{0, x1 - 14}
          s_A(x1) = max{3, x1 + 1}
          f#_A(x1) = max{0, x1 - 15}
          g_A(x1) = max{11, x1 + 1}
          f_A(x1) = max{13, x1}
          minus_A(x1,x2) = max{11, x1, x2 - 15}
          |0|_A = 12
          g#!_A(x1) = max{2, x1 - 14}
          s!_A(x1) = 1
          f#!_A(x1) = max{4, x1 - 12}
          g!_A(x1) = max{4, x1 - 6}
          f!_A(x1) = max{3, x1 - 13}
          minus!_A(x1,x2) = max{2, x1 - 14, x2 - 19}
          |0|!_A = 5
  
    partial status:
  
      pi(g#) = [1]
      pi(s) = [1]
      pi(f#) = [1]
      pi(g) = [1]
      pi(f) = []
      pi(minus) = []
      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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          minus#_A(x1,x2) = x1 + x2 + 4
          s_A(x1) = x1 + 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          minus#_A(x1,x2) = max{x1, x2}
          s_A(x1) = max{15, x1 + 10}
          minus#!_A(x1,x2) = max{3, x1 - 16, x2 - 16}
          s!_A(x1) = max{2, x1 - 2}
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.