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:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      cond1#_A(x1,x2,x3,x4) = x3 + 10
      true_A = 15
      cond2#_A(x1,x2,x3,x4) = max{x1 + 1, x3 + 10}
      gr_A(x1,x2) = max{1, x1 - 4}
      false_A = 0
      p_A(x1) = max{5, x1 - 6}
      |0|_A = 5
      s_A(x1) = max{21, x1 + 20}
    
    2. max/plus interpretations on natural numbers:
    
      cond1#_A(x1,x2,x3,x4) = 0
      true_A = 0
      cond2#_A(x1,x2,x3,x4) = 0
      gr_A(x1,x2) = 1
      false_A = 1
      p_A(x1) = 1
      |0|_A = 0
      s_A(x1) = 0
    

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- 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:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      cond1#_A(x1,x2,x3,x4) = max{x1 - 3, x2 + 1}
      true_A = 5
      cond2#_A(x1,x2,x3,x4) = max{1, x2 - 1}
      gr_A(x1,x2) = max{4, x1 + 1}
      false_A = 0
      p_A(x1) = max{0, x1 - 3}
      |0|_A = 0
      s_A(x1) = max{6, x1 + 5}
    
    2. max/plus interpretations on natural numbers:
    
      cond1#_A(x1,x2,x3,x4) = 0
      true_A = 0
      cond2#_A(x1,x2,x3,x4) = 1
      gr_A(x1,x2) = 1
      false_A = 0
      p_A(x1) = 0
      |0|_A = 0
      s_A(x1) = 0
    

The next rules are strictly ordered:

  p1, p2

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: 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:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      gr#_A(x1,x2) = max{x1, x2}
      s_A(x1) = max{2, x1 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      gr#_A(x1,x2) = 0
      s_A(x1) = 0
    

The next rules are strictly ordered:

  p1

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