YES
We show the termination of the TRS R:

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

and R consists of:

r1: cond(true(),x,y) -> cond(gr(x,y),p(x),s(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{8, x1 - 20, x2}
            true_A = 19
            gr_A(x1,x2) = max{2, x1 - 3}
            p_A(x1) = max{8, x1 - 2}
            s_A(x1) = max{22, x1 + 2}
            |0|_A = 4
            false_A = 2
            cond#!_A(x1,x2,x3) = max{21, x1 + 9, x2 + 12}
            true!_A = 25
            gr!_A(x1,x2) = 24
            p!_A(x1) = max{27, x1 - 4}
            s!_A(x1) = x1 + 29
            |0|!_A = 26
            false!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              cond#_A(x1,x2,x3) = 0
              true_A() = 1
              gr_A(x1,x2) = x2 + 1
              p_A(x1) = x1
              s_A(x1) = x1 + 1
              |0|_A() = 1
              false_A() = 0
              cond#!_A(x1,x2,x3) = x2 + 5
              true!_A() = 4
              gr!_A(x1,x2) = 3
              p!_A(x1) = 0
              s!_A(x1) = 2
              |0|!_A() = 5
              false!_A() = 0
        
  
    partial status:
  
      pi(cond#) = []
      pi(true) = []
      pi(gr) = [1, 2]
      pi(p) = [1]
      pi(s) = [1]
      pi(|0|) = []
      pi(false) = []

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),s(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{8, x1}
            gr#!_A(x1,x2) = max{6, x1 - 9, x2 - 9}
            s!_A(x1) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              gr#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              gr#!_A(x1,x2) = x1 + x2 + 5
              s!_A(x1) = 4
        
  
    partial status:
  
      pi(gr#) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.