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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cond#_A(x1,x2,x3,x4) = max{x1 - 10, x2 + 8, x3 + 11}
          true_A = 22
          and_A(x1,x2) = max{20, x1 + 1, x2 + 4}
          gr_A(x1,x2) = max{13, x1 + 8}
          p_A(x1) = max{0, x1 - 1}
          false_A = 0
          |0|_A = 0
          s_A(x1) = x1 + 23
    
      precedence:
      
        cond# = and = gr = p > true = false = |0| = s
      
      partial status:
    
        pi(cond#) = [2]
        pi(true) = []
        pi(and) = [1, 2]
        pi(gr) = [1]
        pi(p) = []
        pi(false) = []
        pi(|0|) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cond#_A(x1,x2,x3,x4) = 0
          true_A = 4
          and_A(x1,x2) = max{x1 + 14, x2 + 14}
          gr_A(x1,x2) = x1 + 11
          p_A(x1) = 2
          false_A = 13
          |0|_A = 3
          s_A(x1) = x1 + 5
    
      precedence:
      
        |0| > cond# = true = and = gr = p = false = s
      
      partial status:
    
        pi(cond#) = []
        pi(true) = []
        pi(and) = [1, 2]
        pi(gr) = []
        pi(p) = []
        pi(false) = []
        pi(|0|) = []
        pi(s) = [1]
    

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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          gr#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        gr# = s
      
      partial status:
    
        pi(gr#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          gr#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        gr# = s
      
      partial status:
    
        pi(gr#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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