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:

  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, x3 - 1}
            true_A = 2
            gr_A(x1,x2) = max{0, x1 - 5, x2 - 1}
            p_A(x1) = x1 + 10
            |0|_A = 2
            false_A = 0
            s_A(x1) = max{7, x1 + 6}
            cond#!_A(x1,x2,x3) = 1
            true!_A = 5
            gr!_A(x1,x2) = max{7, x1 - 8, x2 + 2}
            p!_A(x1) = max{7, x1 - 3}
            |0|!_A = 8
            false!_A = 3
            s!_A(x1) = max{6, x1 - 7}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond#_A(x1,x2,x3) = max{0, x3 - 1}
            true_A = 10
            gr_A(x1,x2) = max{2, x1 + 1}
            p_A(x1) = max{4, x1 - 1}
            |0|_A = 0
            false_A = 2
            s_A(x1) = max{9, x1 + 1}
            cond#!_A(x1,x2,x3) = max{x1 - 3, x2 + 2}
            true!_A = 9
            gr!_A(x1,x2) = max{11, x1 - 11, x2 - 10}
            p!_A(x1) = max{8, x1 - 1}
            |0|!_A = 10
            false!_A = 0
            s!_A(x1) = max{10, x1 - 10}
        
  
    partial status:
  
      pi(cond#) = []
      pi(true) = []
      pi(gr) = [1]
      pi(p) = [1]
      pi(|0|) = []
      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(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:

  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{4, x1 + 1}
            gr#!_A(x1,x2) = max{3, x1 - 10, x2 - 5}
            s!_A(x1) = max{2, x1 - 11}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gr#_A(x1,x2) = max{x1, x2}
            s_A(x1) = x1 + 1
            gr#!_A(x1,x2) = max{4, x1 - 2, x2 - 2}
            s!_A(x1) = max{3, x1 - 9}
        
  
    partial status:
  
      pi(gr#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.