YES

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

  R:
  g(c(x,s(y))) -> g(c(s(x),y))
  f(c(s(x),y)) -> f(c(x,s(y)))
  f(f(x)) -> f(d(f(x)))
  f(x) -> 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: g#(c(x,s(y))) -> g#(c(s(x),y))
p2: f#(c(s(x),y)) -> f#(c(x,s(y)))
p3: f#(f(x)) -> f#(d(f(x)))

and R consists of:

r1: g(c(x,s(y))) -> g(c(s(x),y))
r2: f(c(s(x),y)) -> f(c(x,s(y)))
r3: f(f(x)) -> f(d(f(x)))
r4: f(x) -> x
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2}


-- Reduction pair.

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

p1: g#(c(x,s(y))) -> g#(c(s(x),y))

and R consists of:

r1: g(c(x,s(y))) -> g(c(s(x),y))
r2: f(c(s(x),y)) -> f(c(x,s(y)))
r3: f(f(x)) -> f(d(f(x)))
r4: f(x) -> x
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

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)
            c_A(x1,x2) = x2 + (1,3)
            s_A(x1) = ((1,0),(1,1)) x1 + (0,5)
            g_A(x1) = (0,0)
            f_A(x1) = ((1,0),(0,0)) x1 + (1,2)
            d_A(x1) = ((0,0),(1,0)) x1 + (1,0)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        g# = c = s = g = f = d = rand
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(s) = []
        pi(g) = []
        pi(f) = []
        pi(d) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = ((1,0),(1,0)) x1
            c_A(x1,x2) = ((1,0),(0,0)) x2 + (1,5)
            s_A(x1) = (2,4)
            g_A(x1) = (0,0)
            f_A(x1) = (4,6)
            d_A(x1) = (0,0)
            rand_A(x1) = (0,0)
    
      precedence:
      
        rand > c = s = f = d > g# = g
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(s) = []
        pi(g) = []
        pi(f) = []
        pi(d) = []
        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: f#(c(s(x),y)) -> f#(c(x,s(y)))

and R consists of:

r1: g(c(x,s(y))) -> g(c(s(x),y))
r2: f(c(s(x),y)) -> f(c(x,s(y)))
r3: f(f(x)) -> f(d(f(x)))
r4: f(x) -> x
r5: rand(x) -> x
r6: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

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

The next rules are strictly ordered:

  p1

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