YES
We show the termination of the TRS R:

  cond1(true(),x,y,z) -> cond2(gr(y,z),x,y,z)
  cond2(true(),x,y,z) -> cond2(gr(y,z),x,p(y),z)
  cond2(false(),x,y,z) -> cond1(gr(x,z),p(x),y,z)
  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: cond1#(true(),x,y,z) -> cond2#(gr(y,z),x,y,z)
p2: cond1#(true(),x,y,z) -> gr#(y,z)
p3: cond2#(true(),x,y,z) -> cond2#(gr(y,z),x,p(y),z)
p4: cond2#(true(),x,y,z) -> gr#(y,z)
p5: cond2#(true(),x,y,z) -> p#(y)
p6: cond2#(false(),x,y,z) -> cond1#(gr(x,z),p(x),y,z)
p7: cond2#(false(),x,y,z) -> gr#(x,z)
p8: cond2#(false(),x,y,z) -> p#(x)
p9: gr#(s(x),s(y)) -> gr#(x,y)

and R consists of:

r1: cond1(true(),x,y,z) -> cond2(gr(y,z),x,y,z)
r2: cond2(true(),x,y,z) -> cond2(gr(y,z),x,p(y),z)
r3: cond2(false(),x,y,z) -> cond1(gr(x,z),p(x),y,z)
r4: gr(|0|(),x) -> false()
r5: gr(s(x),|0|()) -> true()
r6: gr(s(x),s(y)) -> gr(x,y)
r7: p(|0|()) -> |0|()
r8: p(s(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1, p3, p6}
  {p9}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cond1#(true(),x,y,z) -> cond2#(gr(y,z),x,y,z)
p2: cond2#(false(),x,y,z) -> cond1#(gr(x,z),p(x),y,z)
p3: cond2#(true(),x,y,z) -> cond2#(gr(y,z),x,p(y),z)

and R consists of:

r1: cond1(true(),x,y,z) -> cond2(gr(y,z),x,y,z)
r2: cond2(true(),x,y,z) -> cond2(gr(y,z),x,p(y),z)
r3: cond2(false(),x,y,z) -> cond1(gr(x,z),p(x),y,z)
r4: gr(|0|(),x) -> false()
r5: gr(s(x),|0|()) -> true()
r6: gr(s(x),s(y)) -> gr(x,y)
r7: p(|0|()) -> |0|()
r8: p(s(x)) -> x

The set of usable rules consists of

  r4, r5, r6, r7, r8

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:
          
            cond1#_A(x1,x2,x3,x4) = max{0, x1 - 26, x2 - 17, x3 - 17, x4 - 14}
            true_A = 25
            cond2#_A(x1,x2,x3,x4) = max{0, x1 - 27, x2 - 17, x3 - 17, x4 - 14}
            gr_A(x1,x2) = max{14, x1 - 34}
            false_A = 14
            p_A(x1) = max{16, x1 - 17}
            |0|_A = 16
            s_A(x1) = max{59, x1 + 33}
            cond1#!_A(x1,x2,x3,x4) = max{x1 + 14, x2 + 12, x3 - 8, x4 + 9}
            true!_A = 52
            cond2#!_A(x1,x2,x3,x4) = max{x1 + 25, x2 + 11, x3 - 8, x4 + 9}
            gr!_A(x1,x2) = max{51, x2 + 37}
            false!_A = 36
            p!_A(x1) = max{38, x1 - 17}
            |0|!_A = 53
            s!_A(x1) = max{53, x1 - 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond1#_A(x1,x2,x3,x4) = max{0, x1 - 33, x2 - 34, x3 - 34, x4 - 18}
            true_A = 15
            cond2#_A(x1,x2,x3,x4) = max{0, x1 - 17, x2 - 34, x3 - 34, x4 - 18}
            gr_A(x1,x2) = max{0, x1 - 31}
            false_A = 2
            p_A(x1) = max{33, x1 - 34}
            |0|_A = 33
            s_A(x1) = max{46, x1 + 34}
            cond1#!_A(x1,x2,x3,x4) = max{x1 + 30, x2 + 8}
            true!_A = 0
            cond2#!_A(x1,x2,x3,x4) = max{44, x2 + 7}
            gr!_A(x1,x2) = 44
            false!_A = 42
            p!_A(x1) = max{6, x1 - 34}
            |0|!_A = 43
            s!_A(x1) = max{0, x1 - 1}
        
  
    partial status:
  
      pi(cond1#) = [4]
      pi(true) = []
      pi(cond2#) = [4]
      pi(gr) = []
      pi(false) = []
      pi(p) = [1]
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p1, p2



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: cond2#(true(),x,y,z) -> cond2#(gr(y,z),x,p(y),z)

and R consists of:

r1: cond1(true(),x,y,z) -> cond2(gr(y,z),x,y,z)
r2: cond2(true(),x,y,z) -> cond2(gr(y,z),x,p(y),z)
r3: cond2(false(),x,y,z) -> cond1(gr(x,z),p(x),y,z)
r4: gr(|0|(),x) -> false()
r5: gr(s(x),|0|()) -> true()
r6: gr(s(x),s(y)) -> gr(x,y)
r7: p(|0|()) -> |0|()
r8: p(s(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cond2#(true(),x,y,z) -> cond2#(gr(y,z),x,p(y),z)

and R consists of:

r1: cond1(true(),x,y,z) -> cond2(gr(y,z),x,y,z)
r2: cond2(true(),x,y,z) -> cond2(gr(y,z),x,p(y),z)
r3: cond2(false(),x,y,z) -> cond1(gr(x,z),p(x),y,z)
r4: gr(|0|(),x) -> false()
r5: gr(s(x),|0|()) -> true()
r6: gr(s(x),s(y)) -> gr(x,y)
r7: p(|0|()) -> |0|()
r8: p(s(x)) -> x

The set of usable rules consists of

  r4, r5, r6, r7, r8

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:
          
            cond2#_A(x1,x2,x3,x4) = max{0, x1 - 4, x2 - 3, x3 - 9, x4 - 3}
            true_A = 2
            gr_A(x1,x2) = max{2, x1 - 9, x2 - 5}
            p_A(x1) = max{8, x1 - 1}
            |0|_A = 0
            false_A = 0
            s_A(x1) = max{8, x1 + 7}
            cond2#!_A(x1,x2,x3,x4) = max{x1 + 10, x2 + 11, x3 + 2, x4 + 11}
            true!_A = 14
            gr!_A(x1,x2) = 13
            p!_A(x1) = max{15, x1 - 12}
            |0|!_A = 16
            false!_A = 14
            s!_A(x1) = x1 + 14
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond2#_A(x1,x2,x3,x4) = max{0, x1 - 22, x2 - 11, x3 - 11, x4 - 11}
            true_A = 21
            gr_A(x1,x2) = max{12, x1 - 17}
            p_A(x1) = max{5, x1 - 12}
            |0|_A = 0
            false_A = 1
            s_A(x1) = max{38, x1 + 16}
            cond2#!_A(x1,x2,x3,x4) = max{7, x1 - 13, x3 + 1}
            true!_A = 15
            gr!_A(x1,x2) = max{13, x1 - 2}
            p!_A(x1) = max{9, x1 - 32}
            |0|!_A = 16
            false!_A = 14
            s!_A(x1) = max{16, x1 - 1}
        
  
    partial status:
  
      pi(cond2#) = [2]
      pi(true) = []
      pi(gr) = [1, 2]
      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: cond1(true(),x,y,z) -> cond2(gr(y,z),x,y,z)
r2: cond2(true(),x,y,z) -> cond2(gr(y,z),x,p(y),z)
r3: cond2(false(),x,y,z) -> cond1(gr(x,z),p(x),y,z)
r4: gr(|0|(),x) -> false()
r5: gr(s(x),|0|()) -> true()
r6: gr(s(x),s(y)) -> gr(x,y)
r7: p(|0|()) -> |0|()
r8: 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{5, x1}
            gr#!_A(x1,x2) = max{3, x1 - 7, x2 - 6}
            s!_A(x1) = max{2, x1 - 7}
        
        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}
            gr#!_A(x1,x2) = max{3, x1 - 15, x2 - 15}
            s!_A(x1) = max{2, x1 - 14}
        
  
    partial status:
  
      pi(gr#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.