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) = x1 + 10
        if#_A(x1,x2,x3) = max{9, x1 - 10, x2 + 7}
        nonZero_A(x1) = max{9, x1 - 4}
        true_A = 27
        p_A(x1) = max{6, x1 - 4}
        id_inc_A(x1) = max{5, x1 - 1}
        |0|_A = 3
        false_A = 0
        s_A(x1) = x1 + 32
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            rand#_A(x1,x2) = 0
            if#_A(x1,x2,x3) = 0
            nonZero_A(x1) = 1
            true_A() = 1
            p_A(x1) = 2
            id_inc_A(x1) = x1 + 2
            |0|_A() = 1
            false_A() = 1
            s_A(x1) = 1
            rand#!_A(x1,x2) = 3
            if#!_A(x1,x2,x3) = x2 + 7
            nonZero!_A(x1) = 5
            true!_A() = 4
            p!_A(x1) = 6
            id_inc!_A(x1) = 1
            |0|!_A() = 1
            false!_A() = 0
            s!_A(x1) = 5
  
    partial status:
  
      pi(rand#) = []
      pi(if#) = []
      pi(nonZero) = []
      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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        p#_A(x1) = 0
        s_A(x1) = max{0, x1 - 2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            p#_A(x1) = x1
            s_A(x1) = x1 + 2
            p#!_A(x1) = x1 + 1
            s!_A(x1) = 2
  
    partial status:
  
      pi(p#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.