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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        g#_A(x1) = max{1, x1 - 13}
        s_A(x1) = max{24, x1 + 15}
        f#_A(x1) = max{11, x1 - 22}
        g_A(x1) = max{21, x1}
        f_A(x1) = max{24, x1 + 5}
        minus_A(x1,x2) = max{23, x1, x2 - 25}
        |0|_A = 0
  
    base order 2:
  
      precedence:
      
        |0| > f > g > s > g# = f# = minus
      
  
    partial status:
  
      pi(g#) = []
      pi(s) = [1]
      pi(f#) = []
      pi(g) = [1]
      pi(f) = []
      pi(minus) = [1]
      pi(|0|) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: g#(s(x)) -> f#(g(x))
p2: f#(s(x)) -> f#(x)
p3: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: g#(s(x)) -> f#(g(x))
p2: f#(s(x)) -> g#(f(x))
p3: f#(s(x)) -> 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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        g#_A(x1) = max{9, x1 + 3}
        s_A(x1) = max{24, x1 + 8}
        f#_A(x1) = max{10, x1 + 5}
        g_A(x1) = max{18, x1 + 2}
        f_A(x1) = x1 + 9
        minus_A(x1,x2) = max{26, x1 + 1, x2 - 28}
        |0|_A = 17
  
    base order 2:
  
      precedence:
      
        f# > g# > s > f > g > minus = |0|
      
  
    partial status:
  
      pi(g#) = [1]
      pi(s) = [1]
      pi(f#) = [1]
      pi(g) = []
      pi(f) = []
      pi(minus) = [1]
      pi(|0|) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: f#(s(x)) -> g#(f(x))
p2: f#(s(x)) -> 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 estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: f#(s(x)) -> 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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1) = x1 + 1
        s_A(x1) = max{6, x1 + 2}
  
    base order 2:
  
      precedence:
      
        s > f#
      
  
    partial status:
  
      pi(f#) = [1]
      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: 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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        minus#_A(x1,x2) = max{x1 - 1, x2 + 3}
        s_A(x1) = max{5, x1 + 3}
  
    base order 2:
  
      precedence:
      
        minus# > s
      
  
    partial status:
  
      pi(minus#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.