YES

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

  R:
  f(|0|()) -> true()
  f(|1|()) -> false()
  f(s(x)) -> f(x)
  if(true(),s(x),s(y)) -> s(x)
  if(false(),s(x),s(y)) -> s(y)
  g(x,c(y)) -> c(g(x,y))
  g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(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#(s(x)) -> f#(x)
p2: g#(x,c(y)) -> g#(x,y)
p3: g#(x,c(y)) -> g#(x,if(f(x),c(g(s(x),y)),c(y)))
p4: g#(x,c(y)) -> if#(f(x),c(g(s(x),y)),c(y))
p5: g#(x,c(y)) -> f#(x)
p6: g#(x,c(y)) -> g#(s(x),y)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p2, p6}
  {p1}


-- Reduction pair.

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

p1: g#(x,c(y)) -> g#(s(x),y)
p2: g#(x,c(y)) -> g#(x,y)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = max{1, x2 - 4}
          c_A(x1) = x1 + 6
          s_A(x1) = x1
          f_A(x1) = x1
          |0|_A = 2
          true_A = 1
          |1|_A = 2
          false_A = 1
          if_A(x1,x2,x3) = max{x2, x3}
          g_A(x1,x2) = x2
          rand_A(x1) = x1 + 2
    
      precedence:
      
        true > s = f = if > g = rand > c = |1| = false > g# = |0|
      
      partial status:
    
        pi(g#) = []
        pi(c) = [1]
        pi(s) = [1]
        pi(f) = [1]
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = [2, 3]
        pi(g) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = 0
          c_A(x1) = x1 + 21
          s_A(x1) = max{19, x1 + 3}
          f_A(x1) = max{20, x1 + 18}
          |0|_A = 20
          true_A = 21
          |1|_A = 22
          false_A = 21
          if_A(x1,x2,x3) = max{17, x2 - 4, x3}
          g_A(x1,x2) = 0
          rand_A(x1) = 1
    
      precedence:
      
        g# = c = f = |0| = true > s = |1| = false = if = g = rand
      
      partial status:
    
        pi(g#) = []
        pi(c) = [1]
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = []
        pi(g) = []
        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#(x,c(y)) -> g#(s(x),y)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: 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: g#(x,c(y)) -> g#(s(x),y)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = max{x1 + 2, x2 + 1}
          c_A(x1) = x1 + 3
          s_A(x1) = 0
          f_A(x1) = 2
          |0|_A = 2
          true_A = 1
          |1|_A = 1
          false_A = 0
          if_A(x1,x2,x3) = max{2, x1 - 8, x3 - 1}
          g_A(x1,x2) = x2 + 10
          rand_A(x1) = x1
    
      precedence:
      
        g# = c = s = f = |0| = true = |1| = false = if = g = rand
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = []
        pi(g) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = 0
          c_A(x1) = max{1, x1}
          s_A(x1) = 0
          f_A(x1) = 7
          |0|_A = 0
          true_A = 1
          |1|_A = 0
          false_A = 1
          if_A(x1,x2,x3) = 3
          g_A(x1,x2) = 1
          rand_A(x1) = x1 + 2
    
      precedence:
      
        g# = c = f = |0| = true = |1| = false = if = g = rand > s
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = []
        pi(g) = []
        pi(rand) = [1]
    

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#(s(x)) -> f#(x)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1
          s_A(x1) = x1
          f_A(x1) = max{3, x1 + 2}
          |0|_A = 3
          true_A = 4
          |1|_A = 5
          false_A = 4
          if_A(x1,x2,x3) = max{3, x1, x2 + 2, x3 + 1}
          g_A(x1,x2) = max{x1 + 27, x2 + 14}
          c_A(x1) = 11
          rand_A(x1) = x1 + 2
    
      precedence:
      
        s > f# = |0| = |1| = false = if = g > f = true > c = rand
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = [1, 2, 3]
        pi(g) = []
        pi(c) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 5
          s_A(x1) = x1 + 5
          f_A(x1) = 6
          |0|_A = 0
          true_A = 7
          |1|_A = 2
          false_A = 7
          if_A(x1,x2,x3) = max{4, x1 - 1, x2 + 2, x3 + 1}
          g_A(x1,x2) = 2
          c_A(x1) = 3
          rand_A(x1) = 1
    
      precedence:
      
        f# = s = f = |0| = true = |1| = false = if = g = c = rand
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = [3]
        pi(g) = []
        pi(c) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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