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, 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{x1 + 55, x2 + 30}
        true_A = 11
        cond2#_A(x1,x2) = max{64, x2 + 10}
        even_A(x1) = max{7, x1 - 132}
        false_A = 8
        neq_A(x1,x2) = max{9, x1 - 45, x2 - 23}
        |0|_A = 30
        p_A(x1) = max{31, x1 - 31}
        div2_A(x1) = max{34, x1 - 31}
        s_A(x1) = x1 + 66
        y_A = 22
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            cond1#_A(x1,x2) = x2
            true_A() = 8
            cond2#_A(x1,x2) = x2
            even_A(x1) = x1 + 1
            false_A() = 8
            neq_A(x1,x2) = 8
            |0|_A() = 1
            p_A(x1) = x1
            div2_A(x1) = x1
            s_A(x1) = x1 + 5
            y_A() = 1
            cond1#!_A(x1,x2) = 6
            true!_A() = 7
            cond2#!_A(x1,x2) = x2 + 6
            even!_A(x1) = x1 + 5
            false!_A() = 2
            neq!_A(x1,x2) = x1 + 3
            |0|!_A() = 4
            p!_A(x1) = 1
            div2!_A(x1) = x1 + 5
            s!_A(x1) = x1 + 4
            y!_A() = 0
  
    partial status:
  
      pi(cond1#) = []
      pi(true) = []
      pi(cond2#) = [2]
      pi(even) = []
      pi(false) = []
      pi(neq) = []
      pi(|0|) = []
      pi(p) = []
      pi(div2) = []
      pi(s) = [1]
      pi(y) = []

The following rules are removed:

  p1, p3



-- SCC decomposition.

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

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

  (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{7, x1 + 5}
        s_A(x1) = x1 + 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            even#_A(x1) = 0
            s_A(x1) = 1
            even#!_A(x1) = 4
            s!_A(x1) = 3
  
    partial status:
  
      pi(even#) = [1]
      pi(s) = []

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 + 5}
        s_A(x1) = x1 + 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            div2#_A(x1) = 0
            s_A(x1) = 1
            div2#!_A(x1) = 4
            s!_A(x1) = 3
  
    partial status:
  
      pi(div2#) = [1]
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.