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

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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond1#_A(x1,x2) = max{1, x2 - 15}
            true_A = 1
            cond2#_A(x1,x2) = max{1, x1 - 15, x2 - 16}
            even_A(x1) = max{1, x1 - 2}
            false_A = 14
            neq_A(x1,x2) = 13
            |0|_A = 0
            p_A(x1) = max{12, x1 - 7}
            div2_A(x1) = max{0, x1 - 15}
            s_A(x1) = max{25, x1 + 15}
            y_A = 67
            cond1#!_A(x1,x2) = x2 + 14
            true!_A = 5
            cond2#!_A(x1,x2) = max{x1 + 13, x2 + 8}
            even!_A(x1) = max{6, x1 - 24}
            false!_A = 2
            neq!_A(x1,x2) = max{9, x1 - 26, x2 - 125}
            |0|!_A = 28
            p!_A(x1) = max{3, x1 - 1}
            div2!_A(x1) = max{13, x1 - 41}
            s!_A(x1) = max{13, x1 - 57}
            y!_A = 11
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond1#_A(x1,x2) = max{25, x2 + 10}
            true_A = 2
            cond2#_A(x1,x2) = max{25, x1, x2 + 10}
            even_A(x1) = x1 + 3
            false_A = 24
            neq_A(x1,x2) = max{2, x2 - 10}
            |0|_A = 12
            p_A(x1) = max{15, x1 - 5}
            div2_A(x1) = x1
            s_A(x1) = x1 + 9
            y_A = 3
            cond1#!_A(x1,x2) = x2 + 6
            true!_A = 3
            cond2#!_A(x1,x2) = max{x1 + 3, x2 + 6}
            even!_A(x1) = max{19, x1 - 13}
            false!_A = 18
            neq!_A(x1,x2) = max{19, x1 + 6, x2 - 13}
            |0|!_A = 28
            p!_A(x1) = max{1, x1 - 13}
            div2!_A(x1) = 28
            s!_A(x1) = 18
            y!_A = 20
        
  
    partial status:
  
      pi(cond1#) = []
      pi(true) = []
      pi(cond2#) = []
      pi(even) = []
      pi(false) = []
      pi(neq) = []
      pi(|0|) = []
      pi(p) = [1]
      pi(div2) = [1]
      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, r6, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond1#_A(x1,x2) = 1
            true_A = 44
            cond2#_A(x1,x2) = 1
            even_A(x1) = max{50, x1 + 33}
            neq_A(x1,x2) = max{3, x1, x2 - 9}
            |0|_A = 10
            div2_A(x1) = max{13, x1 - 24}
            false_A = 2
            s_A(x1) = x1 + 44
            p_A(x1) = max{10, x1 - 25}
            cond1#!_A(x1,x2) = max{x1 - 4, x2 + 25}
            true!_A = 40
            cond2#!_A(x1,x2) = max{39, x2 + 24}
            even!_A(x1) = 0
            neq!_A(x1,x2) = max{23, x2 - 45}
            |0|!_A = 41
            div2!_A(x1) = max{43, x1 + 36}
            false!_A = 1
            s!_A(x1) = 42
            p!_A(x1) = max{42, x1 - 11}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            cond1#_A(x1,x2) = max{14, x1 + 4, x2 + 1}
            true_A = 9
            cond2#_A(x1,x2) = max{14, x2 + 1}
            even_A(x1) = x1 + 7
            neq_A(x1,x2) = max{1, x1 - 3}
            |0|_A = 2
            div2_A(x1) = max{4, x1 - 12}
            false_A = 1
            s_A(x1) = x1 + 16
            p_A(x1) = max{2, x1 - 16}
            cond1#!_A(x1,x2) = max{x1 + 7, x2 + 10}
            true!_A = 18
            cond2#!_A(x1,x2) = max{15, x1 - 7, x2 + 9}
            even!_A(x1) = 17
            neq!_A(x1,x2) = max{x1 + 8, x2 + 19}
            |0|!_A = 19
            div2!_A(x1) = max{45, x1 - 17}
            false!_A = 30
            s!_A(x1) = x1 + 28
            p!_A(x1) = max{46, x1 - 3}
        
  
    partial status:
  
      pi(cond1#) = []
      pi(true) = []
      pi(cond2#) = []
      pi(even) = [1]
      pi(neq) = [1, 2]
      pi(|0|) = []
      pi(div2) = [1]
      pi(false) = []
      pi(s) = [1]
      pi(p) = [1]

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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            even#_A(x1) = 0
            s_A(x1) = max{8, x1 - 8}
            even#!_A(x1) = max{29, x1 - 30}
            s!_A(x1) = max{29, x1 - 22}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            even#_A(x1) = 0
            s_A(x1) = max{4, x1 - 4}
            even#!_A(x1) = 6
            s!_A(x1) = max{6, x1 - 17}
        
  
    partial status:
  
      pi(even#) = [1]
      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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div2#_A(x1) = 0
            s_A(x1) = max{4, x1 - 4}
            div2#!_A(x1) = max{14, x1 - 12}
            s!_A(x1) = max{14, x1 - 7}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div2#_A(x1) = 0
            s_A(x1) = max{4, x1 - 6}
            div2#!_A(x1) = 5
            s!_A(x1) = max{5, x1 - 9}
        
  
    partial status:
  
      pi(div2#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.