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:

  max/plus interpretations on natural numbers:
  
    rand#_A(x1,x2) = max{0, x1 - 4}
    if#_A(x1,x2,x3) = max{0, x1 - 5, x2 - 4}
    nonZero_A(x1) = x1
    true_A = 8
    p_A(x1) = max{2, x1 - 2}
    id_inc_A(x1) = max{8, x1 + 4}
    |0|_A = 3
    false_A = 1
    s_A(x1) = max{8, x1 + 4}

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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(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:

  (no SCCs)

-- 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:

  max/plus interpretations on natural numbers:
  
    p#_A(x1) = max{0, x1 - 1}
    s_A(x1) = x1 + 2

The next rules are strictly ordered:

  p1

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