YES
We show the termination of the TRS R:

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

and R consists of:

r1: nonZero(|0|()) -> false()
r2: nonZero(s(x)) -> true()
r3: p(|0|()) -> |0|()
r4: p(s(x)) -> 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:

  {p2, p4}


-- 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(|0|()) -> |0|()
r4: p(s(x)) -> 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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. 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) = x1 + 1
              true_A() = 1
              p_A(x1) = x1 + 1
              id_inc_A(x1) = 1
              |0|_A() = 1
              false_A() = 2
              s_A(x1) = x1 + 4
              rand#!_A(x1,x2) = 5
              if#!_A(x1,x2,x3) = 5
              nonZero!_A(x1) = x1
              true!_A() = 3
              p!_A(x1) = 4
              id_inc!_A(x1) = 2
              |0|!_A() = 1
              false!_A() = 0
              s!_A(x1) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            rand#_A(x1,x2) = 1
            if#_A(x1,x2,x3) = 1
            nonZero_A(x1) = max{0, x1 - 9}
            true_A = 8
            p_A(x1) = max{7, x1 - 2}
            id_inc_A(x1) = 0
            |0|_A = 1
            false_A = 0
            s_A(x1) = max{17, x1 + 9}
            rand#!_A(x1,x2) = max{5, x1 - 1}
            if#!_A(x1,x2,x3) = max{4, x1 - 2, x2 - 3}
            nonZero!_A(x1) = 3
            true!_A = 4
            p!_A(x1) = max{7, x1 - 4}
            id_inc!_A(x1) = 3
            |0|!_A = 8
            false!_A = 2
            s!_A(x1) = x1 + 4
        
  
    partial status:
  
      pi(rand#) = []
      pi(if#) = []
      pi(nonZero) = []
      pi(true) = []
      pi(p) = [1]
      pi(id_inc) = []
      pi(|0|) = []
      pi(false) = []
      pi(s) = [1]

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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(|0|()) -> |0|()
r4: p(s(x)) -> 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:

  (no SCCs)