YES

We show the termination of the TRS R:

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}
  {p7}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    cond#_A(x1,x2,x3,x4) = max{2, x1 - 5, x2, x3 - 2}
    true_A = 8
    and_A(x1,x2) = max{x1 + 1, x2 + 1}
    gr_A(x1,x2) = max{1, x1}
    p_A(x1) = max{1, x1 - 1}
    false_A = 1
    |0|_A = 0
    s_A(x1) = x1 + 8

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    gr#_A(x1,x2) = max{x1 + 1, x2 + 1}
    s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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