YES

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

  R:
  f(c(s(x),y)) -> f(c(x,s(y)))
  f(c(s(x),s(y))) -> g(c(x,y))
  g(c(x,s(y))) -> g(c(s(x),y))
  g(c(s(x),s(y))) -> f(c(x,y))

  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: f#(c(s(x),y)) -> f#(c(x,s(y)))
p2: f#(c(s(x),s(y))) -> g#(c(x,y))
p3: g#(c(x,s(y))) -> g#(c(s(x),y))
p4: g#(c(s(x),s(y))) -> f#(c(x,y))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- 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)))
p2: f#(c(s(x),s(y))) -> g#(c(x,y))
p3: g#(c(s(x),s(y))) -> f#(c(x,y))
p4: g#(c(x,s(y))) -> g#(c(s(x),y))

and R consists of:

r1: f(c(s(x),y)) -> f(c(x,s(y)))
r2: f(c(s(x),s(y))) -> g(c(x,y))
r3: g(c(x,s(y))) -> g(c(s(x),y))
r4: g(c(s(x),s(y))) -> f(c(x,y))
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. matrix interpretations:
    
      carrier: N^2
      order: lexicographic order
      interpretations:
        f#_A(x1) = x1
        c_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,1)
        s_A(x1) = x1 + (0,1)
        g#_A(x1) = x1 + (0,1)
        f_A(x1) = x1
        g_A(x1) = x1 + (0,1)
        rand_A(x1) = x1 + (1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > g# > c > f > g > f#
      
      argument filter:
    
        pi(f#) = []
        pi(c) = [2]
        pi(s) = [1]
        pi(g#) = [1]
        pi(f) = []
        pi(g) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p2, p3, p4

We remove them from the problem.

-- SCC decomposition.

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- 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: f(c(s(x),y)) -> f(c(x,s(y)))
r2: f(c(s(x),s(y))) -> g(c(x,y))
r3: g(c(x,s(y))) -> g(c(s(x),y))
r4: g(c(s(x),s(y))) -> f(c(x,y))
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. matrix interpretations:
    
      carrier: N^2
      order: lexicographic order
      interpretations:
        f#_A(x1) = x1
        c_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,1)
        s_A(x1) = x1 + (0,1)
        f_A(x1) = x1
        g_A(x1) = x1 + (0,1)
        rand_A(x1) = ((1,0),(1,0)) x1 + (1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > rand > c > g > f > f#
      
      argument filter:
    
        pi(f#) = 1
        pi(c) = [1]
        pi(s) = [1]
        pi(f) = []
        pi(g) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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