YES
We show the termination of the TRS R:

  cond1(true(),x,y) -> cond2(gr(y,|0|()),x,y)
  cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y))
  cond2(false(),x,y) -> cond1(gr(x,|0|()),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: cond1#(true(),x,y) -> cond2#(gr(y,|0|()),x,y)
p2: cond1#(true(),x,y) -> gr#(y,|0|())
p3: cond2#(true(),x,y) -> cond2#(gr(y,|0|()),x,p(y))
p4: cond2#(true(),x,y) -> gr#(y,|0|())
p5: cond2#(true(),x,y) -> p#(y)
p6: cond2#(false(),x,y) -> cond1#(gr(x,|0|()),p(x),y)
p7: cond2#(false(),x,y) -> gr#(x,|0|())
p8: cond2#(false(),x,y) -> p#(x)
p9: gr#(s(x),s(y)) -> gr#(x,y)

and R consists of:

r1: cond1(true(),x,y) -> cond2(gr(y,|0|()),x,y)
r2: cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y))
r3: cond2(false(),x,y) -> cond1(gr(x,|0|()),p(x),y)
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) -> cond2#(gr(y,|0|()),x,y)
p2: cond2#(false(),x,y) -> cond1#(gr(x,|0|()),p(x),y)
p3: cond2#(true(),x,y) -> cond2#(gr(y,|0|()),x,p(y))

and R consists of:

r1: cond1(true(),x,y) -> cond2(gr(y,|0|()),x,y)
r2: cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y))
r3: cond2(false(),x,y) -> cond1(gr(x,|0|()),p(x),y)
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, 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) = max{17, x2, x3 - 20}
            true_A = 5
            cond2#_A(x1,x2,x3) = max{17, x1 - 8, x2 - 22, x3 - 20}
            gr_A(x1,x2) = max{0, x1 - 16}
            |0|_A = 17
            false_A = 1
            p_A(x1) = max{17, x1 - 22}
            s_A(x1) = x1 + 57
            cond1#!_A(x1,x2,x3) = max{x1 + 35, x2 + 21}
            true!_A = 40
            cond2#!_A(x1,x2,x3) = max{39, x2 + 20}
            gr!_A(x1,x2) = max{2, x1 - 16}
            |0|!_A = 4
            false!_A = 3
            p!_A(x1) = 5
            s!_A(x1) = max{0, x1 - 15}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond1#_A(x1,x2,x3) = max{0, x1 - 13, x2 - 21, x3 - 1}
            true_A = 0
            cond2#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 22, x3 - 1}
            gr_A(x1,x2) = max{0, x1 - 9, x2 - 18}
            |0|_A = 0
            false_A = 0
            p_A(x1) = max{0, x1 - 6}
            s_A(x1) = x1 + 6
            cond1#!_A(x1,x2,x3) = max{x1 + 8, x2 - 1}
            true!_A = 5
            cond2#!_A(x1,x2,x3) = max{x1 + 8, x2 + 15, x3 + 7}
            gr!_A(x1,x2) = max{15, x2 + 9}
            |0|!_A = 17
            false!_A = 15
            p!_A(x1) = max{18, x1 + 16}
            s!_A(x1) = x1 + 4
        
  
    partial status:
  
      pi(cond1#) = []
      pi(true) = []
      pi(cond2#) = []
      pi(gr) = [1, 2]
      pi(|0|) = []
      pi(false) = []
      pi(p) = [1]
      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) -> cond2#(gr(y,|0|()),x,p(y))

and R consists of:

r1: cond1(true(),x,y) -> cond2(gr(y,|0|()),x,y)
r2: cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y))
r3: cond2(false(),x,y) -> cond1(gr(x,|0|()),p(x),y)
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) -> cond2#(gr(y,|0|()),x,p(y))

and R consists of:

r1: cond1(true(),x,y) -> cond2(gr(y,|0|()),x,y)
r2: cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y))
r3: cond2(false(),x,y) -> cond1(gr(x,|0|()),p(x),y)
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, 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) = max{x1 - 3, x2 - 1, x3 + 2}
            true_A = 1
            gr_A(x1,x2) = 4
            |0|_A = 0
            p_A(x1) = x1
            false_A = 2
            s_A(x1) = x1
            cond2#!_A(x1,x2,x3) = x2 + 2
            true!_A = 4
            gr!_A(x1,x2) = max{x1 + 12, x2 + 5}
            |0|!_A = 8
            p!_A(x1) = max{9, x1 - 6}
            false!_A = 7
            s!_A(x1) = max{5, x1 - 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond2#_A(x1,x2,x3) = max{0, x2 - 20, x3 - 33}
            true_A = 22
            gr_A(x1,x2) = max{5, x1 - 10}
            |0|_A = 14
            p_A(x1) = max{14, x1 - 11}
            false_A = 5
            s_A(x1) = max{35, x1 + 11}
            cond2#!_A(x1,x2,x3) = max{x1 + 1, x3 + 8}
            true!_A = 28
            gr!_A(x1,x2) = max{26, x1 - 8, x2 + 15}
            |0|!_A = 31
            p!_A(x1) = max{25, x1 + 9}
            false!_A = 30
            s!_A(x1) = x1 + 29
        
  
    partial status:
  
      pi(cond2#) = []
      pi(true) = []
      pi(gr) = [1, 2]
      pi(|0|) = []
      pi(p) = [1]
      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) -> cond2(gr(y,|0|()),x,y)
r2: cond2(true(),x,y) -> cond2(gr(y,|0|()),x,p(y))
r3: cond2(false(),x,y) -> cond1(gr(x,|0|()),p(x),y)
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) = 0
            s_A(x1) = max{0, x1 - 2}
            gr#!_A(x1,x2) = max{0, x1 - 17}
            s!_A(x1) = max{2, x1 - 16}
        
        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
            gr#!_A(x1,x2) = max{3, x1 - 1, x2 - 1}
            s!_A(x1) = max{2, x1 - 26}
        
  
    partial status:
  
      pi(gr#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.