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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cond1#_A(x1,x2,x3) = max{x1 + 4, x2 + 15}
          true_A = 12
          cond2#_A(x1,x2,x3) = max{15, x2 + 14}
          gr_A(x1,x2) = max{x1 + 8, x2 + 6}
          |0|_A = 0
          false_A = 3
          p_A(x1) = max{0, x1 - 2}
          s_A(x1) = max{14, x1 + 13}
    
      precedence:
      
        cond2# = gr = p = s > cond1# > true = |0| = false
      
      partial status:
    
        pi(cond1#) = [1]
        pi(true) = []
        pi(cond2#) = []
        pi(gr) = []
        pi(|0|) = []
        pi(false) = []
        pi(p) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cond1#_A(x1,x2,x3) = x1
          true_A = 7
          cond2#_A(x1,x2,x3) = 3
          gr_A(x1,x2) = 2
          |0|_A = 6
          false_A = 1
          p_A(x1) = 0
          s_A(x1) = 0
    
      precedence:
      
        |0| > cond1# = cond2# = p = s > true > gr > false
      
      partial status:
    
        pi(cond1#) = [1]
        pi(true) = []
        pi(cond2#) = []
        pi(gr) = []
        pi(|0|) = []
        pi(false) = []
        pi(p) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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: 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:

  {p2}


-- 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cond2#_A(x1,x2,x3) = max{12, x1, x3 + 5}
          true_A = 13
          gr_A(x1,x2) = max{x1 + 2, x2 + 3}
          |0|_A = 1
          p_A(x1) = max{6, x1 - 1}
          false_A = 2
          s_A(x1) = x1 + 14
    
      precedence:
      
        cond2# = true = gr = |0| = p = false = s
      
      partial status:
    
        pi(cond2#) = [1, 3]
        pi(true) = []
        pi(gr) = [1, 2]
        pi(|0|) = []
        pi(p) = []
        pi(false) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cond2#_A(x1,x2,x3) = max{x1 + 2, x3 + 3}
          true_A = 1
          gr_A(x1,x2) = max{x1 + 4, x2 + 6}
          |0|_A = 2
          p_A(x1) = 4
          false_A = 5
          s_A(x1) = x1 + 2
    
      precedence:
      
        cond2# = true = gr = |0| = p = false = s
      
      partial status:
    
        pi(cond2#) = [3]
        pi(true) = []
        pi(gr) = [1, 2]
        pi(|0|) = []
        pi(p) = []
        pi(false) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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) -> 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          gr#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        gr# = s
      
      partial status:
    
        pi(gr#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          gr#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        gr# = s
      
      partial status:
    
        pi(gr#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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