YES

We show the termination of the relative TRS R/S:

  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)))

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

Consider the non-minimal 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)))
r7: rand(x) -> x
r8: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

Consider the non-minimal 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)))
r7: rand(x) -> x
r8: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{1, x1}
          s_A(x1) = max{4, x1}
          f#_A(x1) = x1
          g_A(x1) = max{2, x1}
          f_A(x1) = max{4, x1}
          minus_A(x1,x2) = x1
          |0|_A = 3
          rand_A(x1) = max{6, x1 + 1}
    
      precedence:
      
        s = f > f# > g# = g > minus = |0| = rand
      
      partial status:
    
        pi(g#) = [1]
        pi(s) = [1]
        pi(f#) = [1]
        pi(g) = [1]
        pi(f) = [1]
        pi(minus) = [1]
        pi(|0|) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{3, x1}
          s_A(x1) = max{8, x1 + 6}
          f#_A(x1) = max{8, x1 - 5}
          g_A(x1) = x1 + 11
          f_A(x1) = max{8, x1 + 1}
          minus_A(x1,x2) = max{9, x1 + 1}
          |0|_A = 1
          rand_A(x1) = 0
    
      precedence:
      
        g > g# > f# > f > s > minus = |0| = rand
      
      partial status:
    
        pi(g#) = [1]
        pi(s) = [1]
        pi(f#) = []
        pi(g) = []
        pi(f) = [1]
        pi(minus) = [1]
        pi(|0|) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1, p4

We remove them from the problem.

-- SCC decomposition.

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

p1: g#(s(x)) -> f#(g(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)))
r7: rand(x) -> x
r8: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

Consider the non-minimal 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)))
r7: rand(x) -> x
r8: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{1, x1}
          s_A(x1) = max{1, x1}
          minus_A(x1,x2) = max{x1 + 3, x2 - 8}
          |0|_A = 8
          f_A(x1) = x1 + 8
          g_A(x1) = max{6, x1 + 5}
          rand_A(x1) = max{4, x1 + 2}
    
      precedence:
      
        g > minus > f# = s > |0| = f > rand
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(minus) = [1]
        pi(|0|) = []
        pi(f) = [1]
        pi(g) = [1]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{20, x1}
          s_A(x1) = max{31, x1 + 21}
          minus_A(x1,x2) = max{22, x1 + 4}
          |0|_A = 27
          f_A(x1) = max{26, x1 + 2}
          g_A(x1) = max{18, x1 + 4}
          rand_A(x1) = 1
    
      precedence:
      
        g > f# = s = minus = |0| = f = rand
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = []
        pi(minus) = [1]
        pi(|0|) = []
        pi(f) = [1]
        pi(g) = [1]
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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)))
r7: rand(x) -> x
r8: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = max{x1 + 1, x2 + 1}
          s_A(x1) = max{5, x1}
          minus_A(x1,x2) = max{6, x1 + 1}
          |0|_A = 4
          f_A(x1) = max{16, x1 + 9}
          g_A(x1) = x1 + 7
          rand_A(x1) = max{12, x1 + 6}
    
      precedence:
      
        s > |0| = f > minus# = g > minus = rand
      
      partial status:
    
        pi(minus#) = [2]
        pi(s) = [1]
        pi(minus) = [1]
        pi(|0|) = []
        pi(f) = [1]
        pi(g) = [1]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = x2
          s_A(x1) = max{23, x1 + 18}
          minus_A(x1,x2) = max{24, x1 + 2}
          |0|_A = 21
          f_A(x1) = max{22, x1 + 9}
          g_A(x1) = max{9, x1 + 3}
          rand_A(x1) = 1
    
      precedence:
      
        |0| > f > s > g > minus# = minus = rand
      
      partial status:
    
        pi(minus#) = [2]
        pi(s) = []
        pi(minus) = [1]
        pi(|0|) = []
        pi(f) = [1]
        pi(g) = [1]
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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