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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1) = x1 + 2
        s_A(x1) = x1
        f#_A(x1) = x1 + 2
        g_A(x1) = x1
        f_A(x1) = x1
        minus_A(x1,x2) = x1
        |0|_A = 1
        rand_A(x1) = x1 + 2
  
    precedence:
    
      |0| > g# = s = f# = f > g > minus > 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) = []

The next rules are strictly ordered:

  p2

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)) -> 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)))
r7: rand(x) -> x
r8: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}


-- 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = max{1, x1}
        s_A(x1) = max{2, x1}
        minus_A(x1,x2) = max{x1 + 6, x2 - 13}
        |0|_A = 19
        f_A(x1) = max{18, x1 + 12}
        g_A(x1) = x1 + 6
        rand_A(x1) = max{6, x1 + 3}
  
    precedence:
    
      |0| > f# = s = minus = f = g = rand
    
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(f) = []
      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: 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 set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1) = max{1, x1}
        s_A(x1) = max{2, x1}
        minus_A(x1,x2) = max{x1 + 6, x2 - 13}
        |0|_A = 19
        f_A(x1) = max{18, x1 + 12}
        g_A(x1) = x1 + 6
        rand_A(x1) = max{6, x1 + 3}
  
    precedence:
    
      |0| > g# = s = minus = f = g = rand
    
    partial status:
  
      pi(g#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(f) = []
      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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = max{x1 + 1, x2 + 1}
        s_A(x1) = max{4, x1}
        minus_A(x1,x2) = max{5, x1, x2 - 13}
        |0|_A = 3
        f_A(x1) = max{18, x1 + 12}
        g_A(x1) = x1 + 6
        rand_A(x1) = max{10, x1 + 5}
  
    precedence:
    
      minus# = s = minus = |0| = f = g = rand
    
    partial status:
  
      pi(minus#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(f) = []
      pi(g) = [1]
      pi(rand) = []

The next rules are strictly ordered:

  p1

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