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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          cond1#_A(x1,x2,x3,x4) = 1
          true_A() = 0
          cond2#_A(x1,x2,x3,x4) = 1
          gr_A(x1,x2) = x1 + x2 + 2
          false_A() = 3
          p_A(x1) = x1 + 4
          |0|_A() = 4
          s_A(x1) = x1 + 1
  
    base order 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 - 18, x3 - 10, x4 - 34}
          true_A = 15
          cond2#_A(x1,x2,x3,x4) = max{0, x1 - 17, x2 - 18, x3 - 10, x4 - 34}
          gr_A(x1,x2) = x1
          false_A = 2
          p_A(x1) = max{9, x1 - 4}
          |0|_A = 3
          s_A(x1) = x1 + 19
          cond1#!_A(x1,x2,x3,x4) = max{x1 - 1, x2 + 3}
          true!_A = 14
          cond2#!_A(x1,x2,x3,x4) = max{13, x2 + 2}
          gr!_A(x1,x2) = max{18, x2 + 2}
          false!_A = 16
          p!_A(x1) = max{15, x1 - 11}
          |0|!_A = 17
          s!_A(x1) = max{19, x1 + 1}
  
    partial status:
  
      pi(cond1#) = []
      pi(true) = []
      pi(cond2#) = []
      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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          cond2#_A(x1,x2,x3,x4) = x1 + x2 + 1
          true_A() = 2
          gr_A(x1,x2) = 2
          p_A(x1) = x1 + 4
          |0|_A() = 3
          false_A() = 1
          s_A(x1) = x1 + 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          cond2#_A(x1,x2,x3,x4) = max{0, x1 - 55, x2 - 28, x3 - 12, x4 - 28}
          true_A = 53
          gr_A(x1,x2) = max{28, x1 - 6}
          p_A(x1) = max{11, x1 - 4}
          |0|_A = 5
          false_A = 1
          s_A(x1) = max{59, x1 + 4}
          cond2#!_A(x1,x2,x3,x4) = max{x1 - 38, x3 + 3}
          true!_A = 27
          gr!_A(x1,x2) = max{83, x1 + 24, x2 + 23}
          p!_A(x1) = max{1, x1 - 9}
          |0|!_A = 31
          false!_A = 30
          s!_A(x1) = max{84, x1 - 1}
  
    partial status:
  
      pi(cond2#) = [2]
      pi(true) = []
      pi(gr) = []
      pi(p) = []
      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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          gr#_A(x1,x2) = x1 + x2 + 7
          s_A(x1) = x1 + 6
  
    base order 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{3, x1 + 1}
          gr#!_A(x1,x2) = max{5, x1 - 2, x2 - 4}
          s!_A(x1) = 4
  
    partial status:
  
      pi(gr#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.