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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              cond1#_A(x1,x2,x3,x4) = x1
              true_A() = 1
              cond2#_A(x1,x2,x3,x4) = 1
              gr_A(x1,x2) = 1
              false_A() = 1
              p_A(x1) = x1 + 3
              |0|_A() = 3
              s_A(x1) = x1 + 1
              cond1#!_A(x1,x2,x3,x4) = x4 + 1
              true!_A() = 4
              cond2#!_A(x1,x2,x3,x4) = x1 + x4
              gr!_A(x1,x2) = x2 + 1
              false!_A() = 0
              p!_A(x1) = 3
              |0|!_A() = 5
              s!_A(x1) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond1#_A(x1,x2,x3,x4) = max{1, x4 - 1}
            true_A = 20
            cond2#_A(x1,x2,x3,x4) = max{1, x4 - 1}
            gr_A(x1,x2) = x1 + 3
            false_A = 3
            p_A(x1) = max{11, x1 - 5}
            |0|_A = 11
            s_A(x1) = x1 + 17
            cond1#!_A(x1,x2,x3,x4) = max{16, x3 + 4}
            true!_A = 14
            cond2#!_A(x1,x2,x3,x4) = max{16, x1 - 3, x3 + 4}
            gr!_A(x1,x2) = 16
            false!_A = 17
            p!_A(x1) = x1 + 7
            |0|!_A = 18
            s!_A(x1) = 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:

  p3



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


-- 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)

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              cond1#_A(x1,x2,x3,x4) = 0
              true_A() = 1
              cond2#_A(x1,x2,x3,x4) = 0
              gr_A(x1,x2) = 1
              false_A() = 1
              p_A(x1) = x1 + 1
              |0|_A() = 3
              s_A(x1) = x1 + 1
              cond1#!_A(x1,x2,x3,x4) = x3 + x4 + 2
              true!_A() = 3
              cond2#!_A(x1,x2,x3,x4) = x3 + x4 + 2
              gr!_A(x1,x2) = x1 + x2 + 1
              false!_A() = 3
              p!_A(x1) = 5
              |0|!_A() = 4
              s!_A(x1) = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond1#_A(x1,x2,x3,x4) = max{0, x1 - 18, x2 - 20, x3 - 20, x4 - 17}
            true_A = 17
            cond2#_A(x1,x2,x3,x4) = max{0, x1 - 19, x2 - 20, x3 - 20, x4 - 17}
            gr_A(x1,x2) = max{0, x1 - 2}
            false_A = 0
            p_A(x1) = max{4, x1 - 18}
            |0|_A = 0
            s_A(x1) = max{19, x1 + 18}
            cond1#!_A(x1,x2,x3,x4) = max{x1 - 1, x2 + 10}
            true!_A = 0
            cond2#!_A(x1,x2,x3,x4) = max{15, x2 + 9}
            gr!_A(x1,x2) = 7
            false!_A = 1
            p!_A(x1) = max{24, x1 - 1}
            |0|!_A = 2
            s!_A(x1) = max{0, x1 - 20}
        
  
    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

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              gr#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              gr#!_A(x1,x2) = x1 + x2
              s!_A(x1) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gr#_A(x1,x2) = 0
            s_A(x1) = max{1, x1 - 1}
            gr#!_A(x1,x2) = max{10, x2 - 2}
            s!_A(x1) = max{9, x1 - 1}
        
  
    partial status:
  
      pi(gr#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.