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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        cond1#_A(x1,x2,x3) = max{x1 + 7, x2 + 14}
        true_A = 16
        cond2#_A(x1,x2,x3) = max{22, x2 + 14}
        gr_A(x1,x2) = max{12, x1 + 6}
        |0|_A = 7
        false_A = 12
        p_A(x1) = max{7, x1 - 6}
        s_A(x1) = max{13, x1 + 11}
  
    base order 2:
  
      precedence:
      
        cond1# > cond2# > true = false = s > gr = p > |0|
      
  
    partial status:
  
      pi(cond1#) = []
      pi(true) = []
      pi(cond2#) = []
      pi(gr) = []
      pi(|0|) = []
      pi(false) = []
      pi(p) = []
      pi(s) = []

The following rules are removed:

  p2



-- 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, r6, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        cond2#_A(x1,x2,x3) = max{x1 - 6, x3 + 11}
        true_A = 27
        gr_A(x1,x2) = max{5, x1 + 2}
        |0|_A = 8
        p_A(x1) = max{9, x1 - 12}
        false_A = 3
        s_A(x1) = max{28, x1 + 26}
  
    base order 2:
  
      precedence:
      
        cond2# > |0| = s > true > gr > false > p
      
  
    partial status:
  
      pi(cond2#) = []
      pi(true) = []
      pi(gr) = [1]
      pi(|0|) = []
      pi(p) = []
      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

  r4, r5, r6, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        gr#_A(x1,x2) = max{x1 - 1, x2 + 16}
        s_A(x1) = max{21, x1 + 15}
        gr_A(x1,x2) = max{4, x1 - 22, x2 - 16}
        |0|_A = 11
        false_A = 1
        true_A = 3
        p_A(x1) = max{13, x1 - 14}
  
    base order 2:
  
      precedence:
      
        p > |0| > gr# = gr = true > s = false
      
  
    partial status:
  
      pi(gr#) = [2]
      pi(s) = [1]
      pi(gr) = []
      pi(|0|) = []
      pi(false) = []
      pi(true) = []
      pi(p) = []

The following rules are removed:

  p1

Then no dependency pair remains.