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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = x1 + (0,1)
            s_A(x1) = ((1,0),(1,1)) x1 + (0,6)
            f#_A(x1) = x1 + (0,3)
            g_A(x1) = ((1,0),(1,1)) x1 + (0,2)
            f_A(x1) = ((1,0),(1,1)) x1 + (0,7)
            minus_A(x1,x2) = x1 + (0,1)
            |0|_A() = (1,0)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        f = rand > s > f# > g# > g > minus = |0|
      
      partial status:
    
        pi(g#) = [1]
        pi(s) = [1]
        pi(f#) = []
        pi(g) = []
        pi(f) = [1]
        pi(minus) = []
        pi(|0|) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = ((1,0),(1,1)) x1 + (1,2)
            s_A(x1) = ((0,0),(1,0)) x1 + (4,1)
            f#_A(x1) = (3,6)
            g_A(x1) = (0,0)
            f_A(x1) = x1 + (1,2)
            minus_A(x1,x2) = (0,0)
            |0|_A() = (5,7)
            rand_A(x1) = (0,0)
    
      precedence:
      
        s > rand > g = minus > f# = f = |0| > g#
      
      partial status:
    
        pi(g#) = [1]
        pi(s) = []
        pi(f#) = []
        pi(g) = []
        pi(f) = [1]
        pi(minus) = []
        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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = x1 + (1,2)
            s_A(x1) = x1 + (0,1)
            minus_A(x1,x2) = ((1,0),(0,0)) x1 + (1,2)
            |0|_A() = (1,1)
            f_A(x1) = x1 + (2,2)
            g_A(x1) = x1 + (2,2)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,2)
    
      precedence:
      
        s = minus = |0| = f = g = rand > f#
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = []
        pi(minus) = []
        pi(|0|) = []
        pi(f) = []
        pi(g) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = (0,0)
            s_A(x1) = (0,2)
            minus_A(x1,x2) = (0,0)
            |0|_A() = (1,1)
            f_A(x1) = (2,3)
            g_A(x1) = (2,3)
            rand_A(x1) = (0,0)
    
      precedence:
      
        f# = s > minus = |0| = f = g = rand
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
        pi(minus) = []
        pi(|0|) = []
        pi(f) = []
        pi(g) = []
        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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            minus#_A(x1,x2) = x2
            s_A(x1) = x1 + (0,5)
            minus_A(x1,x2) = x1 + (1,1)
            |0|_A() = (1,1)
            f_A(x1) = x1 + (2,2)
            g_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,0)
    
      precedence:
      
        minus# > s = f = g > minus = |0| = rand
      
      partial status:
    
        pi(minus#) = [2]
        pi(s) = [1]
        pi(minus) = [1]
        pi(|0|) = []
        pi(f) = []
        pi(g) = [1]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            minus#_A(x1,x2) = (0,0)
            s_A(x1) = (3,3)
            minus_A(x1,x2) = (1,1)
            |0|_A() = (1,1)
            f_A(x1) = (4,4)
            g_A(x1) = (2,2)
            rand_A(x1) = (0,0)
    
      precedence:
      
        s > minus > rand > minus# = |0| = f = g
      
      partial status:
    
        pi(minus#) = []
        pi(s) = []
        pi(minus) = []
        pi(|0|) = []
        pi(f) = []
        pi(g) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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