YES
We show the termination of the TRS R:

  cond1(true(),x) -> cond2(even(x),x)
  cond2(true(),x) -> cond1(neq(x,|0|()),div2(x))
  cond2(false(),x) -> cond1(neq(x,|0|()),p(x))
  neq(|0|(),|0|()) -> false()
  neq(|0|(),s(x)) -> true()
  neq(s(x),|0|()) -> true()
  neq(s(x),s(y())) -> neq(x,y())
  even(|0|()) -> true()
  even(s(|0|())) -> false()
  even(s(s(x))) -> even(x)
  div2(|0|()) -> |0|()
  div2(s(|0|())) -> |0|()
  div2(s(s(x))) -> s(div2(x))
  p(|0|()) -> |0|()
  p(s(x)) -> x


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: cond1#(true(),x) -> cond2#(even(x),x)
p2: cond1#(true(),x) -> even#(x)
p3: cond2#(true(),x) -> cond1#(neq(x,|0|()),div2(x))
p4: cond2#(true(),x) -> neq#(x,|0|())
p5: cond2#(true(),x) -> div2#(x)
p6: cond2#(false(),x) -> cond1#(neq(x,|0|()),p(x))
p7: cond2#(false(),x) -> neq#(x,|0|())
p8: cond2#(false(),x) -> p#(x)
p9: neq#(s(x),s(y())) -> neq#(x,y())
p10: even#(s(s(x))) -> even#(x)
p11: div2#(s(s(x))) -> div2#(x)

and R consists of:

r1: cond1(true(),x) -> cond2(even(x),x)
r2: cond2(true(),x) -> cond1(neq(x,|0|()),div2(x))
r3: cond2(false(),x) -> cond1(neq(x,|0|()),p(x))
r4: neq(|0|(),|0|()) -> false()
r5: neq(|0|(),s(x)) -> true()
r6: neq(s(x),|0|()) -> true()
r7: neq(s(x),s(y())) -> neq(x,y())
r8: even(|0|()) -> true()
r9: even(s(|0|())) -> false()
r10: even(s(s(x))) -> even(x)
r11: div2(|0|()) -> |0|()
r12: div2(s(|0|())) -> |0|()
r13: div2(s(s(x))) -> s(div2(x))
r14: p(|0|()) -> |0|()
r15: p(s(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1, p3, p6}
  {p10}
  {p11}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cond1#(true(),x) -> cond2#(even(x),x)
p2: cond2#(false(),x) -> cond1#(neq(x,|0|()),p(x))
p3: cond2#(true(),x) -> cond1#(neq(x,|0|()),div2(x))

and R consists of:

r1: cond1(true(),x) -> cond2(even(x),x)
r2: cond2(true(),x) -> cond1(neq(x,|0|()),div2(x))
r3: cond2(false(),x) -> cond1(neq(x,|0|()),p(x))
r4: neq(|0|(),|0|()) -> false()
r5: neq(|0|(),s(x)) -> true()
r6: neq(s(x),|0|()) -> true()
r7: neq(s(x),s(y())) -> neq(x,y())
r8: even(|0|()) -> true()
r9: even(s(|0|())) -> false()
r10: even(s(s(x))) -> even(x)
r11: div2(|0|()) -> |0|()
r12: div2(s(|0|())) -> |0|()
r13: div2(s(s(x))) -> s(div2(x))
r14: p(|0|()) -> |0|()
r15: p(s(x)) -> x

The set of usable rules consists of

  r4, r6, r11, r12, r13, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        cond1#_A(x1,x2) = max{6, x1 - 3, x2 + 5}
        true_A = 19
        cond2#_A(x1,x2) = max{16, x2 - 1}
        even_A(x1) = 4
        false_A = 5
        neq_A(x1,x2) = max{1, x1}
        |0|_A = 8
        p_A(x1) = max{9, x1 - 7}
        div2_A(x1) = max{10, x1 - 7}
        s_A(x1) = x1 + 20
  
    base order 2:
  
      precedence:
      
        cond1# > cond2# = s > |0| = p = div2 > true > even = neq > false
      
  
    partial status:
  
      pi(cond1#) = []
      pi(true) = []
      pi(cond2#) = []
      pi(even) = []
      pi(false) = []
      pi(neq) = [1]
      pi(|0|) = []
      pi(p) = []
      pi(div2) = []
      pi(s) = []

The following rules are removed:

  p1



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: cond2#(false(),x) -> cond1#(neq(x,|0|()),p(x))
p2: cond2#(true(),x) -> cond1#(neq(x,|0|()),div2(x))

and R consists of:

r1: cond1(true(),x) -> cond2(even(x),x)
r2: cond2(true(),x) -> cond1(neq(x,|0|()),div2(x))
r3: cond2(false(),x) -> cond1(neq(x,|0|()),p(x))
r4: neq(|0|(),|0|()) -> false()
r5: neq(|0|(),s(x)) -> true()
r6: neq(s(x),|0|()) -> true()
r7: neq(s(x),s(y())) -> neq(x,y())
r8: even(|0|()) -> true()
r9: even(s(|0|())) -> false()
r10: even(s(s(x))) -> even(x)
r11: div2(|0|()) -> |0|()
r12: div2(s(|0|())) -> |0|()
r13: div2(s(s(x))) -> s(div2(x))
r14: p(|0|()) -> |0|()
r15: p(s(x)) -> x

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: even#(s(s(x))) -> even#(x)

and R consists of:

r1: cond1(true(),x) -> cond2(even(x),x)
r2: cond2(true(),x) -> cond1(neq(x,|0|()),div2(x))
r3: cond2(false(),x) -> cond1(neq(x,|0|()),p(x))
r4: neq(|0|(),|0|()) -> false()
r5: neq(|0|(),s(x)) -> true()
r6: neq(s(x),|0|()) -> true()
r7: neq(s(x),s(y())) -> neq(x,y())
r8: even(|0|()) -> true()
r9: even(s(|0|())) -> false()
r10: even(s(s(x))) -> even(x)
r11: div2(|0|()) -> |0|()
r12: div2(s(|0|())) -> |0|()
r13: div2(s(s(x))) -> s(div2(x))
r14: p(|0|()) -> |0|()
r15: p(s(x)) -> x

The set of usable rules consists of

  r5, r6, r7, r8, r9, r11, r12, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        even#_A(x1) = x1 + 6
        s_A(x1) = max{33, x1 + 7}
        neq_A(x1,x2) = max{16, x1 - 34, x2 - 16}
        |0|_A = 29
        true_A = 14
        y_A = 8
        even_A(x1) = max{41, x1 - 38}
        false_A = 1
        div2_A(x1) = max{32, x1 - 41}
        p_A(x1) = max{35, x1 - 6}
  
    base order 2:
  
      precedence:
      
        even# = neq = even > s > div2 = p > |0| = y > true = false
      
  
    partial status:
  
      pi(even#) = [1]
      pi(s) = [1]
      pi(neq) = []
      pi(|0|) = []
      pi(true) = []
      pi(y) = []
      pi(even) = []
      pi(false) = []
      pi(div2) = []
      pi(p) = []

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: div2#(s(s(x))) -> div2#(x)

and R consists of:

r1: cond1(true(),x) -> cond2(even(x),x)
r2: cond2(true(),x) -> cond1(neq(x,|0|()),div2(x))
r3: cond2(false(),x) -> cond1(neq(x,|0|()),p(x))
r4: neq(|0|(),|0|()) -> false()
r5: neq(|0|(),s(x)) -> true()
r6: neq(s(x),|0|()) -> true()
r7: neq(s(x),s(y())) -> neq(x,y())
r8: even(|0|()) -> true()
r9: even(s(|0|())) -> false()
r10: even(s(s(x))) -> even(x)
r11: div2(|0|()) -> |0|()
r12: div2(s(|0|())) -> |0|()
r13: div2(s(s(x))) -> s(div2(x))
r14: p(|0|()) -> |0|()
r15: p(s(x)) -> x

The set of usable rules consists of

  r5, r6, r7, r8, r9, r11, r12, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div2#_A(x1) = x1 + 5
        s_A(x1) = max{31, x1 + 6}
        neq_A(x1,x2) = max{14, x1 - 32, x2 - 14}
        |0|_A = 26
        true_A = 13
        y_A = 7
        even_A(x1) = max{38, x1 - 33}
        false_A = 15
        div2_A(x1) = max{30, x1 - 38}
        p_A(x1) = max{30, x1 - 5}
  
    base order 2:
  
      precedence:
      
        div2# = neq = even > s > div2 = p > |0| = y > true = false
      
  
    partial status:
  
      pi(div2#) = [1]
      pi(s) = [1]
      pi(neq) = []
      pi(|0|) = []
      pi(true) = []
      pi(y) = []
      pi(even) = []
      pi(false) = []
      pi(div2) = []
      pi(p) = []

The following rules are removed:

  p1

Then no dependency pair remains.