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, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        cond1#_A(x1,x2) = max{50, x1 + 5, x2 - 12}
        true_A = 47
        cond2#_A(x1,x2) = max{52, x1 - 62, x2 - 20}
        even_A(x1) = max{49, x1 - 49}
        false_A = 35
        neq_A(x1,x2) = max{47, x1 - 25, x2 - 31}
        |0|_A = 10
        p_A(x1) = max{36, x1 - 23}
        div2_A(x1) = max{11, x1 - 9}
        s_A(x1) = x1 + 24
        y_A = 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          cond1#_A(x1,x2) = max{0, x1 - 10, x2 - 18}
          true_A = 0
          cond2#_A(x1,x2) = max{0, x1 - 9, x2 - 28}
          even_A(x1) = max{0, x1 - 28}
          false_A = 7
          neq_A(x1,x2) = max{9, x1 - 30, x2 - 28}
          |0|_A = 17
          p_A(x1) = max{17, x1 - 18}
          div2_A(x1) = max{17, x1 - 10}
          s_A(x1) = max{27, x1 + 18}
          y_A = 26
          cond1#!_A(x1,x2) = max{0, x2 - 29}
          true!_A = 1
          cond2#!_A(x1,x2) = max{0, x1 - 1, x2 - 32}
          even!_A(x1) = max{8, x1 - 37}
          false!_A = 4
          neq!_A(x1,x2) = max{2, x2 - 83}
          |0|!_A = 1
          p!_A(x1) = max{5, x1 - 68}
          div2!_A(x1) = max{2, x1 - 103}
          s!_A(x1) = max{3, x1 - 56}
          y!_A = 3
  
    partial status:
  
      pi(cond1#) = []
      pi(true) = []
      pi(cond2#) = []
      pi(even) = []
      pi(false) = []
      pi(neq) = []
      pi(|0|) = []
      pi(p) = []
      pi(div2) = []
      pi(s) = [1]
      pi(y) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: cond1#(true(),x) -> cond2#(even(x),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:

  {p1, p2}


-- Reduction pair.

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

p1: cond1#(true(),x) -> cond2#(even(x),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 set of usable rules consists of

  r4, r5, r6, r11, r12, r13

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        cond1#_A(x1,x2) = max{41, x1 + 3, x2 + 27}
        true_A = 40
        cond2#_A(x1,x2) = max{42, x2 + 26}
        even_A(x1) = 20
        neq_A(x1,x2) = max{38, x1 + 22, x2 + 16}
        |0|_A = 11
        div2_A(x1) = max{12, x1 - 10}
        false_A = 37
        s_A(x1) = max{41, x1 + 21}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          cond1#_A(x1,x2) = max{46, x1 - 50, x2 + 7}
          true_A = 49
          cond2#_A(x1,x2) = max{46, x2 - 11}
          even_A(x1) = 61
          neq_A(x1,x2) = max{48, x1 + 39, x2 + 30}
          |0|_A = 10
          div2_A(x1) = max{39, x1 - 18}
          false_A = 40
          s_A(x1) = 18
          cond1#!_A(x1,x2) = max{20, x1 - 20, x2 + 18}
          true!_A = 3
          cond2#!_A(x1,x2) = 19
          even!_A(x1) = 2
          neq!_A(x1,x2) = max{2, x2 - 11}
          |0|!_A = 4
          div2!_A(x1) = max{20, x1 - 19}
          false!_A = 6
          s!_A(x1) = 5
  
    partial status:
  
      pi(cond1#) = []
      pi(true) = []
      pi(cond2#) = [2]
      pi(even) = []
      pi(neq) = [2]
      pi(|0|) = []
      pi(div2) = []
      pi(false) = []
      pi(s) = []

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: 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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        even#_A(x1) = max{1, x1 - 11}
        s_A(x1) = max{13, x1 + 5}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          even#_A(x1) = 0
          s_A(x1) = max{9, x1 - 4}
          even#!_A(x1) = max{5, x1 - 10}
          s!_A(x1) = max{4, x1 - 11}
  
    partial status:
  
      pi(even#) = []
      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: 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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div2#_A(x1) = max{7, x1 - 7}
        s_A(x1) = max{15, x1 + 3}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div2#_A(x1) = 0
          s_A(x1) = max{10, x1 - 4}
          div2#!_A(x1) = max{9, x1 - 34}
          s!_A(x1) = max{8, x1 - 33}
  
    partial status:
  
      pi(div2#) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.