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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        f > s > g# > f# > g > |0| > minus
      
      argument filter:
    
        pi(g#) = 1
        pi(s) = 1
        pi(f#) = 1
        pi(g) = 1
        pi(f) = 1
        pi(minus) = 1
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        g#_A(x1) = x1
        s_A(x1) = x1 + 5
        f#_A(x1) = x1 + 2
        g_A(x1) = x1 + 2
        f_A(x1) = x1 + 6
        minus_A(x1,x2) = x1 + 1
        |0|_A() = 1
    

The next rules are strictly ordered:

  p1, p2, p3, p4

We remove them from the problem.  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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > minus#
      
      argument filter:
    
        pi(minus#) = 2
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        minus#_A(x1,x2) = 0
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.