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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = x1 + x2 + (4,3)
            c_A(x1) = ((1,0),(0,0)) x1 + (3,2)
            s_A(x1) = (0,0)
            f_A(x1) = (2,0)
            |0|_A() = (2,1)
            true_A() = (1,0)
            |1|_A() = (2,2)
            false_A() = (1,1)
            if_A(x1,x2,x3) = x2
            g_A(x1,x2) = x2 + (0,1)
            rand_A(x1) = x1 + (1,1)
    
      precedence:
      
        true = |1| = g > c = f > |0| = if > s = false > g# = rand
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = []
        pi(g) = [2]
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = (0,0)
            c_A(x1) = (2,4)
            s_A(x1) = (0,0)
            f_A(x1) = (0,3)
            |0|_A() = (1,1)
            true_A() = (0,2)
            |1|_A() = (1,2)
            false_A() = (0,1)
            if_A(x1,x2,x3) = (1,6)
            g_A(x1,x2) = (3,5)
            rand_A(x1) = (1,1)
    
      precedence:
      
        rand > s = |0| = true = if = g > |1| = false > g# > c = f
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = []
        pi(g) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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#(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#(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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (7,1)
            c_A(x1) = ((1,0),(0,0)) x1 + (6,0)
            f_A(x1) = (3,7)
            |0|_A() = (2,1)
            true_A() = (1,2)
            |1|_A() = (1,1)
            false_A() = (2,2)
            s_A(x1) = (0,0)
            if_A(x1,x2,x3) = ((1,0),(0,0)) x1 + (1,1)
            g_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (1,0)
            rand_A(x1) = x1 + (1,1)
    
      precedence:
      
        g# = c = f = |0| = true = |1| = false = s = if = g = rand
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(s) = []
        pi(if) = []
        pi(g) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = x1 + x2
            c_A(x1) = (0,1)
            f_A(x1) = (2,0)
            |0|_A() = (2,5)
            true_A() = (1,4)
            |1|_A() = (2,2)
            false_A() = (1,1)
            s_A(x1) = (3,3)
            if_A(x1,x2,x3) = (0,0)
            g_A(x1,x2) = x1 + (0,2)
            rand_A(x1) = (0,0)
    
      precedence:
      
        if = g > f = |1| = false = s > g# > rand > c > |0| = true
      
      partial status:
    
        pi(g#) = [1, 2]
        pi(c) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(s) = []
        pi(if) = []
        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: 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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = x1
            s_A(x1) = ((1,0),(1,1)) x1 + (0,2)
            f_A(x1) = ((1,0),(1,1)) x1 + (2,1)
            |0|_A() = (2,1)
            true_A() = (1,0)
            |1|_A() = (0,1)
            false_A() = (2,0)
            if_A(x1,x2,x3) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + x3 + (1,1)
            g_A(x1,x2) = (6,6)
            c_A(x1) = (1,3)
            rand_A(x1) = ((1,0),(0,0)) x1 + (1,3)
    
      precedence:
      
        |0| = if > s > f# > f = true = false > g > |1| = c = rand
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(f) = [1]
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = [3]
        pi(g) = []
        pi(c) = []
        pi(rand) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = x1 + (1,1)
            s_A(x1) = ((1,0),(1,1)) x1 + (1,1)
            f_A(x1) = ((1,0),(1,0)) x1 + (5,5)
            |0|_A() = (3,10)
            true_A() = (2,9)
            |1|_A() = (1,8)
            false_A() = (0,7)
            if_A(x1,x2,x3) = x3 + (1,1)
            g_A(x1,x2) = (4,4)
            c_A(x1) = (2,2)
            rand_A(x1) = (0,0)
    
      precedence:
      
        if > s > |0| = true > f = false = g = c > f# > |1| > rand
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = []
        pi(g) = []
        pi(c) = []
        pi(rand) = []
    

The next rules are strictly ordered:

  p1

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