YES

We show the termination of the TRS R:

  cond(true(),x) -> cond(and(even(x),gr(x,|0|())),p(x))
  and(x,false()) -> false()
  and(false(),x) -> false()
  and(true(),true()) -> true()
  even(|0|()) -> true()
  even(s(|0|())) -> false()
  even(s(s(x))) -> even(x)
  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) -> cond#(and(even(x),gr(x,|0|())),p(x))
p2: cond#(true(),x) -> and#(even(x),gr(x,|0|()))
p3: cond#(true(),x) -> even#(x)
p4: cond#(true(),x) -> gr#(x,|0|())
p5: cond#(true(),x) -> p#(x)
p6: even#(s(s(x))) -> even#(x)
p7: gr#(s(x),s(y())) -> gr#(x,y())

and R consists of:

r1: cond(true(),x) -> cond(and(even(x),gr(x,|0|())),p(x))
r2: and(x,false()) -> false()
r3: and(false(),x) -> false()
r4: and(true(),true()) -> true()
r5: even(|0|()) -> true()
r6: even(s(|0|())) -> false()
r7: even(s(s(x))) -> even(x)
r8: gr(|0|(),x) -> false()
r9: gr(s(x),|0|()) -> true()
r10: gr(s(x),s(y())) -> gr(x,y())
r11: p(|0|()) -> |0|()
r12: p(s(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1}
  {p6}


-- Reduction pair.

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

p1: cond#(true(),x) -> cond#(and(even(x),gr(x,|0|())),p(x))

and R consists of:

r1: cond(true(),x) -> cond(and(even(x),gr(x,|0|())),p(x))
r2: and(x,false()) -> false()
r3: and(false(),x) -> false()
r4: and(true(),true()) -> true()
r5: even(|0|()) -> true()
r6: even(s(|0|())) -> false()
r7: even(s(s(x))) -> even(x)
r8: gr(|0|(),x) -> false()
r9: gr(s(x),|0|()) -> true()
r10: gr(s(x),s(y())) -> gr(x,y())
r11: p(|0|()) -> |0|()
r12: p(s(x)) -> x

The set of usable rules consists of

  r2, r3, r4, r5, r6, r7, r8, r9, r11, r12

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      cond#_A(x1,x2) = max{0, x1 - 7, x2 - 2}
      true_A = 8
      and_A(x1,x2) = max{4, x1 - 6, x2 + 2}
      even_A(x1) = max{10, x1 - 5}
      gr_A(x1,x2) = max{x1 + 2, x2 + 2}
      |0|_A = 0
      p_A(x1) = max{1, x1 - 1}
      false_A = 1
      s_A(x1) = max{8, x1 + 7}
    
    2. max/plus interpretations on natural numbers:
    
      cond#_A(x1,x2) = 0
      true_A = 2
      and_A(x1,x2) = 1
      even_A(x1) = 3
      gr_A(x1,x2) = 3
      |0|_A = 0
      p_A(x1) = 1
      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: even#(s(s(x))) -> even#(x)

and R consists of:

r1: cond(true(),x) -> cond(and(even(x),gr(x,|0|())),p(x))
r2: and(x,false()) -> false()
r3: and(false(),x) -> false()
r4: and(true(),true()) -> true()
r5: even(|0|()) -> true()
r6: even(s(|0|())) -> false()
r7: even(s(s(x))) -> even(x)
r8: gr(|0|(),x) -> false()
r9: gr(s(x),|0|()) -> true()
r10: gr(s(x),s(y())) -> gr(x,y())
r11: p(|0|()) -> |0|()
r12: 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:
    
      even#_A(x1) = x1
      s_A(x1) = max{2, x1 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      even#_A(x1) = max{0, x1 - 2}
      s_A(x1) = 0
    

The next rules are strictly ordered:

  p1

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