YES

We show the termination of the TRS R:

  cond(true(),x,y) -> cond(gr(x,y),p(x),y)
  gr(|0|(),x) -> false()
  gr(s(x),|0|()) -> true()
  gr(s(x),s(y)) -> gr(x,y)
  p(|0|()) -> |0|()
  p(s(x)) -> x

-- SCC decomposition.

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

p1: cond#(true(),x,y) -> cond#(gr(x,y),p(x),y)
p2: cond#(true(),x,y) -> gr#(x,y)
p3: cond#(true(),x,y) -> p#(x)
p4: gr#(s(x),s(y)) -> gr#(x,y)

and R consists of:

r1: cond(true(),x,y) -> cond(gr(x,y),p(x),y)
r2: gr(|0|(),x) -> false()
r3: gr(s(x),|0|()) -> true()
r4: gr(s(x),s(y)) -> gr(x,y)
r5: p(|0|()) -> |0|()
r6: p(s(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1}
  {p4}


-- Reduction pair.

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

p1: cond#(true(),x,y) -> cond#(gr(x,y),p(x),y)

and R consists of:

r1: cond(true(),x,y) -> cond(gr(x,y),p(x),y)
r2: gr(|0|(),x) -> false()
r3: gr(s(x),|0|()) -> true()
r4: gr(s(x),s(y)) -> gr(x,y)
r5: p(|0|()) -> |0|()
r6: p(s(x)) -> x

The set of usable rules consists of

  r2, r3, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      cond#_A(x1,x2,x3) = max{9, x1 - 1, x2}
      true_A = 11
      gr_A(x1,x2) = x1
      p_A(x1) = max{8, x1 - 1}
      |0|_A = 7
      false_A = 6
      s_A(x1) = max{13, x1 + 12}
    
    2. max/plus interpretations on natural numbers:
    
      cond#_A(x1,x2,x3) = 0
      true_A = 2
      gr_A(x1,x2) = 1
      p_A(x1) = 1
      |0|_A = 0
      false_A = 0
      s_A(x1) = 0
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: gr#(s(x),s(y)) -> gr#(x,y)

and R consists of:

r1: cond(true(),x,y) -> cond(gr(x,y),p(x),y)
r2: gr(|0|(),x) -> false()
r3: gr(s(x),|0|()) -> true()
r4: gr(s(x),s(y)) -> gr(x,y)
r5: p(|0|()) -> |0|()
r6: p(s(x)) -> x

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      gr#_A(x1,x2) = max{x1, x2}
      s_A(x1) = max{2, x1 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      gr#_A(x1,x2) = 0
      s_A(x1) = 0
    

The next rules are strictly ordered:

  p1

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