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, r10, r11, r12

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) = max{5, x2 - 1}
            true_A = 39
            and_A(x1,x2) = max{6, x2}
            even_A(x1) = x1 + 38
            gr_A(x1,x2) = max{x1 + 31, x2 - 1}
            |0|_A = 6
            p_A(x1) = max{6, x1 - 8}
            false_A = 6
            s_A(x1) = x1 + 8
            y_A = 0
            cond#!_A(x1,x2) = max{x1 - 7, x2 + 25}
            true!_A = 22
            and!_A(x1,x2) = max{20, x1 - 18, x2 + 15}
            even!_A(x1) = max{17, x1 + 1}
            gr!_A(x1,x2) = max{x1 - 9, x2 + 27}
            |0|!_A = 22
            p!_A(x1) = max{23, x1 - 7}
            false!_A = 16
            s!_A(x1) = x1 + 23
            y!_A = 23
        
        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) = 0
              true_A() = 1
              and_A(x1,x2) = x1 + 1
              even_A(x1) = x1 + 1
              gr_A(x1,x2) = 1
              |0|_A() = 1
              p_A(x1) = x1 + 1
              false_A() = 1
              s_A(x1) = x1
              y_A() = 1
              cond#!_A(x1,x2) = 6
              true!_A() = 2
              and!_A(x1,x2) = 7
              even!_A(x1) = x1
              gr!_A(x1,x2) = 5
              |0|!_A() = 2
              p!_A(x1) = 3
              false!_A() = 0
              s!_A(x1) = 3
              y!_A() = 0
        
  
    partial status:
  
      pi(cond#) = []
      pi(true) = []
      pi(and) = [2]
      pi(even) = [1]
      pi(gr) = [1]
      pi(|0|) = []
      pi(p) = [1]
      pi(false) = []
      pi(s) = [1]
      pi(y) = []

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: 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:

  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:
          
            even#_A(x1) = x1
            s_A(x1) = max{10, x1 + 5}
            even#!_A(x1) = max{1, x1 - 14}
            s!_A(x1) = max{7, x1 - 8}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              even#_A(x1) = x1
              s_A(x1) = x1 + 5
              even#!_A(x1) = x1 + 5
              s!_A(x1) = 6
        
  
    partial status:
  
      pi(even#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.