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:
  
      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, x1 - 80, x2 - 32}
            true_A = 45
            cond2#_A(x1,x2) = max{1, x2 - 32}
            even_A(x1) = x1 + 14
            false_A = 48
            neq_A(x1,x2) = max{48, x1 - 32, x2 - 59}
            |0|_A = 31
            p_A(x1) = max{33, x1 - 32}
            div2_A(x1) = max{31, x1 - 9}
            s_A(x1) = x1 + 58
            y_A = 17
            cond1#!_A(x1,x2) = max{30, x1 - 58, x2}
            true!_A = 25
            cond2#!_A(x1,x2) = max{29, x1 - 14, x2 - 9}
            even!_A(x1) = max{9, x1 - 5}
            false!_A = 8
            neq!_A(x1,x2) = max{28, x1 - 32, x2 - 5}
            |0|!_A = 24
            p!_A(x1) = max{23, x1 - 8}
            div2!_A(x1) = 9
            s!_A(x1) = max{28, x1 + 10}
            y!_A = 11
        
        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) = 0
              true_A() = 0
              cond2#_A(x1,x2) = 0
              even_A(x1) = x1 + 1
              false_A() = 8
              neq_A(x1,x2) = 8
              |0|_A() = 7
              p_A(x1) = x1
              div2_A(x1) = 7
              s_A(x1) = x1
              y_A() = 1
              cond1#!_A(x1,x2) = 7
              true!_A() = 12
              cond2#!_A(x1,x2) = 7
              even!_A(x1) = 6
              false!_A() = 5
              neq!_A(x1,x2) = 1
              |0|!_A() = 8
              p!_A(x1) = 9
              div2!_A(x1) = 9
              s!_A(x1) = 7
              y!_A() = 0
        
  
    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

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, x1 - 27, x2 - 8}
            true_A = 25
            cond2#_A(x1,x2) = max{1, x2 - 16}
            even_A(x1) = 25
            neq_A(x1,x2) = max{12, x1 + 11}
            |0|_A = 3
            div2_A(x1) = max{6, x1 - 8}
            false_A = 14
            s_A(x1) = x1 + 14
            cond1#!_A(x1,x2) = max{x1 - 13, x2 + 5}
            true!_A = 13
            cond2#!_A(x1,x2) = max{12, x2 + 5}
            even!_A(x1) = x1 + 10
            neq!_A(x1,x2) = x2 + 31
            |0|!_A = 17
            div2!_A(x1) = max{34, x1 + 30}
            false!_A = 1
            s!_A(x1) = x1 + 53
        
        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) = 0
              true_A() = 6
              cond2#_A(x1,x2) = 0
              even_A(x1) = x1 + 2
              neq_A(x1,x2) = x1 + 1
              |0|_A() = 4
              div2_A(x1) = x1 + 1
              false_A() = 1
              s_A(x1) = x1 + 5
              cond1#!_A(x1,x2) = x2 + 4
              true!_A() = 6
              cond2#!_A(x1,x2) = x1 + 1
              even!_A(x1) = x1 + 3
              neq!_A(x1,x2) = 3
              |0|!_A() = 5
              div2!_A(x1) = 8
              false!_A() = 3
              s!_A(x1) = 4
        
  
    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]

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{13, x1 + 11}
            even#!_A(x1) = max{7, x1 - 14}
            s!_A(x1) = x1 + 6
        
        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) = x1
              s_A(x1) = x1 + 2
              even#!_A(x1) = 3
              s!_A(x1) = 4
        
  
    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

  r7, r9

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{25, x1 - 28}
            neq_A(x1,x2) = max{18, x2 - 1}
            y_A = 0
            even_A(x1) = max{21, x1 - 26}
            |0|_A = 17
            false_A = 19
            div2#!_A(x1) = x1 + 58
            s!_A(x1) = x1 + 87
            neq!_A(x1,x2) = max{15, x2 - 12}
            y!_A = 14
            even!_A(x1) = max{102, x1 - 26}
            |0|!_A = 3
            false!_A = 103
        
        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) = x1
              s_A(x1) = x1 + 2
              neq_A(x1,x2) = x1 + 1
              y_A() = 1
              even_A(x1) = x1 + 1
              |0|_A() = 2
              false_A() = 5
              div2#!_A(x1) = 7
              s!_A(x1) = 8
              neq!_A(x1,x2) = 1
              y!_A() = 0
              even!_A(x1) = x1 + 1
              |0|!_A() = 7
              false!_A() = 6
        
  
    partial status:
  
      pi(div2#) = [1]
      pi(s) = [1]
      pi(neq) = []
      pi(y) = []
      pi(even) = [1]
      pi(|0|) = []
      pi(false) = []

The following rules are removed:

  p1

Then no dependency pair remains.