YES
We show the termination of the TRS R:

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

and R consists of:

r1: cond(true(),x,y) -> cond(and(gr(x,|0|()),gr(y,|0|())),p(x),p(y))
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) -> cond#(and(gr(x,|0|()),gr(y,|0|())),p(x),p(y))

and R consists of:

r1: cond(true(),x,y) -> cond(and(gr(x,|0|()),gr(y,|0|())),p(x),p(y))
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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond#_A(x1,x2,x3) = max{0, x1 - 16, x2 - 6, x3 - 17}
            true_A = 14
            and_A(x1,x2) = x2 + 9
            gr_A(x1,x2) = max{3, x1 - 20, x2 - 49}
            |0|_A = 5
            p_A(x1) = max{5, x1 - 20}
            false_A = 1
            s_A(x1) = max{34, x1 + 20}
            cond#!_A(x1,x2,x3) = max{25, x1 + 12, x2 + 19, x3 + 13}
            true!_A = 2
            and!_A(x1,x2) = max{49, x1 - 11, x2 - 4}
            gr!_A(x1,x2) = max{49, x1 - 35, x2 - 1}
            |0|!_A = 2
            p!_A(x1) = max{0, x1 - 36}
            false!_A = 3
            s!_A(x1) = max{3, x1 - 34}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond#_A(x1,x2,x3) = max{0, x1 - 7, x2 - 10, x3 - 10}
            true_A = 0
            and_A(x1,x2) = max{3, x1 - 1, x2 - 1}
            gr_A(x1,x2) = max{2, x1 - 6, x2 - 1}
            |0|_A = 0
            p_A(x1) = max{9, x1 - 1}
            false_A = 0
            s_A(x1) = x1 + 5
            cond#!_A(x1,x2,x3) = max{4, x1 - 14, x2 - 17, x3 - 10}
            true!_A = 2
            and!_A(x1,x2) = max{7, x1 - 20, x2 - 17}
            gr!_A(x1,x2) = max{5, x1 + 1, x2 - 35}
            |0|!_A = 2
            p!_A(x1) = max{3, x1 - 25}
            false!_A = 6
            s!_A(x1) = max{1, x1 - 36}
        
  
    partial status:
  
      pi(cond#) = [2]
      pi(true) = []
      pi(and) = [2]
      pi(gr) = [1]
      pi(|0|) = []
      pi(p) = [1]
      pi(false) = []
      pi(s) = [1]

The following rules are removed:

  p1

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(and(gr(x,|0|()),gr(y,|0|())),p(x),p(y))
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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gr#_A(x1,x2) = max{x1, x2}
            s_A(x1) = max{2, x1}
            gr#!_A(x1,x2) = max{4, x1 - 14, x2 - 3}
            s!_A(x1) = max{3, x1 - 13}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gr#_A(x1,x2) = max{x1, x2}
            s_A(x1) = max{2, x1 + 1}
            gr#!_A(x1,x2) = max{3, x1 - 16, x2 - 3}
            s!_A(x1) = max{2, x1 - 15}
        
  
    partial status:
  
      pi(gr#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.