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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        rand#_A(x1,x2) = x1 + 11
        if#_A(x1,x2,x3) = max{10, x1 + 3, x2 + 8}
        nonZero_A(x1) = max{3, x1 - 2}
        true_A = 18
        p_A(x1) = max{9, x1 - 16}
        id_inc_A(x1) = max{22, x1 - 25}
        |0|_A = 1
        false_A = 0
        s_A(x1) = max{23, x1 + 17}
  
    base order 2:
  
      precedence:
      
        id_inc > nonZero = s > true > if# = p > rand# = |0| > false
      
  
    partial status:
  
      pi(rand#) = []
      pi(if#) = [1]
      pi(nonZero) = []
      pi(true) = []
      pi(p) = []
      pi(id_inc) = []
      pi(|0|) = []
      pi(false) = []
      pi(s) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: rand#(x,y) -> if#(nonZero(x),x,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)