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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        rand#_A(x1,x2) = max{30, x1 + 15}
        if#_A(x1,x2,x3) = max{29, x1 - 32, x2 + 14}
        nonZero_A(x1) = x1 + 46
        true_A = 63
        p_A(x1) = max{13, x1 - 5}
        id_inc_A(x1) = 64
        |0|_A = 2
        false_A = 1
        s_A(x1) = max{18, x1 + 6}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          rand#_A(x1,x2) = max{1, x1 - 1}
          if#_A(x1,x2,x3) = max{1, x1 - 15, x2 - 1}
          nonZero_A(x1) = x1 + 14
          true_A = 12
          p_A(x1) = 2
          id_inc_A(x1) = max{3, x1}
          |0|_A = 1
          false_A = 1
          s_A(x1) = 1
          rand#!_A(x1,x2) = max{0, x1 - 3}
          if#!_A(x1,x2,x3) = max{0, x2 - 4}
          nonZero!_A(x1) = max{4, x1 - 6}
          true!_A = 4
          p!_A(x1) = max{5, x1 - 12}
          id_inc!_A(x1) = 7
          |0|!_A = 2
          false!_A = 0
          s!_A(x1) = 3
  
    partial status:
  
      pi(rand#) = []
      pi(if#) = []
      pi(nonZero) = [1]
      pi(true) = []
      pi(p) = []
      pi(id_inc) = []
      pi(|0|) = []
      pi(false) = []
      pi(s) = []

The following rules are removed:

  p1, p2

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

  r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        p#_A(x1) = max{17, x1}
        s_A(x1) = max{11, x1 + 6}
        nonZero_A(x1) = max{2, x1 - 12}
        true_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          p#_A(x1) = 0
          s_A(x1) = max{4, x1 - 4}
          nonZero_A(x1) = x1 + 4
          true_A = 1
          p#!_A(x1) = x1 + 10
          s!_A(x1) = x1 + 13
          nonZero!_A(x1) = max{11, x1 - 5}
          true!_A = 4
  
    partial status:
  
      pi(p#) = [1]
      pi(s) = [1]
      pi(nonZero) = []
      pi(true) = []

The following rules are removed:

  p1

Then no dependency pair remains.