YES

We show the termination of the TRS R:

  nonZero(|0|()) -> false()
  nonZero(s(x)) -> true()
  p(s(|0|())) -> |0|()
  p(s(s(x))) -> s(p(s(x)))
  id_inc(x) -> x
  id_inc(x) -> s(x)
  random(x) -> rand(x,|0|())
  rand(x,y) -> if(nonZero(x),x,y)
  if(false(),x,y) -> y
  if(true(),x,y) -> rand(p(x),id_inc(y))

-- SCC decomposition.

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

p1: p#(s(s(x))) -> p#(s(x))
p2: random#(x) -> rand#(x,|0|())
p3: rand#(x,y) -> if#(nonZero(x),x,y)
p4: rand#(x,y) -> nonZero#(x)
p5: if#(true(),x,y) -> rand#(p(x),id_inc(y))
p6: if#(true(),x,y) -> p#(x)
p7: if#(true(),x,y) -> id_inc#(y)

and R consists of:

r1: nonZero(|0|()) -> false()
r2: nonZero(s(x)) -> true()
r3: p(s(|0|())) -> |0|()
r4: p(s(s(x))) -> s(p(s(x)))
r5: id_inc(x) -> x
r6: id_inc(x) -> s(x)
r7: random(x) -> rand(x,|0|())
r8: rand(x,y) -> if(nonZero(x),x,y)
r9: if(false(),x,y) -> y
r10: if(true(),x,y) -> rand(p(x),id_inc(y))

The estimated dependency graph contains the following SCCs:

  {p3, p5}
  {p1}


-- Reduction pair.

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

p1: rand#(x,y) -> if#(nonZero(x),x,y)
p2: if#(true(),x,y) -> rand#(p(x),id_inc(y))

and R consists of:

r1: nonZero(|0|()) -> false()
r2: nonZero(s(x)) -> true()
r3: p(s(|0|())) -> |0|()
r4: p(s(s(x))) -> s(p(s(x)))
r5: id_inc(x) -> x
r6: id_inc(x) -> s(x)
r7: random(x) -> rand(x,|0|())
r8: rand(x,y) -> if(nonZero(x),x,y)
r9: if(false(),x,y) -> y
r10: if(true(),x,y) -> rand(p(x),id_inc(y))

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:
    
        max/plus interpretations on natural numbers:
        
          rand#_A(x1,x2) = x1 + 6
          if#_A(x1,x2,x3) = max{x1 + 1, x2 + 6}
          nonZero_A(x1) = max{5, x1}
          true_A = 8
          p_A(x1) = max{0, x1 - 1}
          id_inc_A(x1) = x1 + 10
          |0|_A = 3
          false_A = 4
          s_A(x1) = x1 + 9
    
      precedence:
      
        nonZero = p = false > rand# > |0| > true = s > if# = id_inc
      
      partial status:
    
        pi(rand#) = []
        pi(if#) = [1]
        pi(nonZero) = [1]
        pi(true) = []
        pi(p) = []
        pi(id_inc) = [1]
        pi(|0|) = []
        pi(false) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          rand#_A(x1,x2) = 2
          if#_A(x1,x2,x3) = x1 + 2
          nonZero_A(x1) = x1 + 9
          true_A = 8
          p_A(x1) = 5
          id_inc_A(x1) = max{2, x1 + 1}
          |0|_A = 4
          false_A = 4
          s_A(x1) = max{3, x1}
    
      precedence:
      
        |0| > rand# = if# = nonZero = true = p = false > s > id_inc
      
      partial status:
    
        pi(rand#) = []
        pi(if#) = [1]
        pi(nonZero) = [1]
        pi(true) = []
        pi(p) = []
        pi(id_inc) = [1]
        pi(|0|) = []
        pi(false) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: p#(s(s(x))) -> p#(s(x))

and R consists of:

r1: nonZero(|0|()) -> false()
r2: nonZero(s(x)) -> true()
r3: p(s(|0|())) -> |0|()
r4: p(s(s(x))) -> s(p(s(x)))
r5: id_inc(x) -> x
r6: id_inc(x) -> s(x)
r7: random(x) -> rand(x,|0|())
r8: rand(x,y) -> if(nonZero(x),x,y)
r9: if(false(),x,y) -> y
r10: if(true(),x,y) -> rand(p(x),id_inc(y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1) = x1 + 1
          s_A(x1) = x1
    
      precedence:
      
        p# = s
      
      partial status:
    
        pi(p#) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1) = x1 + 3
          s_A(x1) = x1 + 1
    
      precedence:
      
        p# > s
      
      partial status:
    
        pi(p#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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