YES
We show the termination of the TRS R:

  check(|0|()) -> zero()
  check(s(|0|())) -> odd()
  check(s(s(|0|()))) -> even()
  check(s(s(s(x)))) -> check(s(x))
  half(|0|()) -> |0|()
  half(s(|0|())) -> |0|()
  half(s(s(x))) -> s(half(x))
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))
  times(x,y) -> timesIter(x,y,|0|())
  timesIter(x,y,z) -> if(check(x),x,y,z,plus(z,y))
  p(s(x)) -> x
  p(|0|()) -> |0|()
  if(zero(),x,y,z,u) -> z
  if(odd(),x,y,z,u) -> timesIter(p(x),y,u)
  if(even(),x,y,z,u) -> plus(timesIter(half(x),y,half(z)),timesIter(half(x),y,half(s(z))))


-- SCC decomposition.

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

p1: check#(s(s(s(x)))) -> check#(s(x))
p2: half#(s(s(x))) -> half#(x)
p3: plus#(s(x),y) -> plus#(x,y)
p4: times#(x,y) -> timesIter#(x,y,|0|())
p5: timesIter#(x,y,z) -> if#(check(x),x,y,z,plus(z,y))
p6: timesIter#(x,y,z) -> check#(x)
p7: timesIter#(x,y,z) -> plus#(z,y)
p8: if#(odd(),x,y,z,u) -> timesIter#(p(x),y,u)
p9: if#(odd(),x,y,z,u) -> p#(x)
p10: if#(even(),x,y,z,u) -> plus#(timesIter(half(x),y,half(z)),timesIter(half(x),y,half(s(z))))
p11: if#(even(),x,y,z,u) -> timesIter#(half(x),y,half(z))
p12: if#(even(),x,y,z,u) -> half#(x)
p13: if#(even(),x,y,z,u) -> half#(z)
p14: if#(even(),x,y,z,u) -> timesIter#(half(x),y,half(s(z)))
p15: if#(even(),x,y,z,u) -> half#(s(z))

and R consists of:

r1: check(|0|()) -> zero()
r2: check(s(|0|())) -> odd()
r3: check(s(s(|0|()))) -> even()
r4: check(s(s(s(x)))) -> check(s(x))
r5: half(|0|()) -> |0|()
r6: half(s(|0|())) -> |0|()
r7: half(s(s(x))) -> s(half(x))
r8: plus(|0|(),y) -> y
r9: plus(s(x),y) -> s(plus(x,y))
r10: times(x,y) -> timesIter(x,y,|0|())
r11: timesIter(x,y,z) -> if(check(x),x,y,z,plus(z,y))
r12: p(s(x)) -> x
r13: p(|0|()) -> |0|()
r14: if(zero(),x,y,z,u) -> z
r15: if(odd(),x,y,z,u) -> timesIter(p(x),y,u)
r16: if(even(),x,y,z,u) -> plus(timesIter(half(x),y,half(z)),timesIter(half(x),y,half(s(z))))

The estimated dependency graph contains the following SCCs:

  {p5, p8, p11, p14}
  {p1}
  {p2}
  {p3}


-- Reduction pair.

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

p1: if#(even(),x,y,z,u) -> timesIter#(half(x),y,half(s(z)))
p2: timesIter#(x,y,z) -> if#(check(x),x,y,z,plus(z,y))
p3: if#(even(),x,y,z,u) -> timesIter#(half(x),y,half(z))
p4: if#(odd(),x,y,z,u) -> timesIter#(p(x),y,u)

and R consists of:

r1: check(|0|()) -> zero()
r2: check(s(|0|())) -> odd()
r3: check(s(s(|0|()))) -> even()
r4: check(s(s(s(x)))) -> check(s(x))
r5: half(|0|()) -> |0|()
r6: half(s(|0|())) -> |0|()
r7: half(s(s(x))) -> s(half(x))
r8: plus(|0|(),y) -> y
r9: plus(s(x),y) -> s(plus(x,y))
r10: times(x,y) -> timesIter(x,y,|0|())
r11: timesIter(x,y,z) -> if(check(x),x,y,z,plus(z,y))
r12: p(s(x)) -> x
r13: p(|0|()) -> |0|()
r14: if(zero(),x,y,z,u) -> z
r15: if(odd(),x,y,z,u) -> timesIter(p(x),y,u)
r16: if(even(),x,y,z,u) -> plus(timesIter(half(x),y,half(z)),timesIter(half(x),y,half(s(z))))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r12, r13

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if#_A(x1,x2,x3,x4,x5) = max{x1 - 81, x2 + 2}
        even_A = 88
        timesIter#_A(x1,x2,x3) = x1 + 3
        half_A(x1) = max{3, x1 - 13}
        s_A(x1) = max{48, x1 + 17}
        check_A(x1) = max{83, x1 + 41}
        plus_A(x1,x2) = max{31, x1 - 43}
        odd_A = 88
        p_A(x1) = max{3, x1 - 4}
        |0|_A = 2
        zero_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if#_A(x1,x2,x3,x4,x5) = max{1, x3 - 1}
          even_A = 4
          timesIter#_A(x1,x2,x3) = max{1, x2 - 1}
          half_A(x1) = max{4, x1}
          s_A(x1) = x1 + 14
          check_A(x1) = 27
          plus_A(x1,x2) = x1 + 3
          odd_A = 1
          p_A(x1) = max{8, x1 - 6}
          |0|_A = 5
          zero_A = 1
          if#!_A(x1,x2,x3,x4,x5) = max{x1 + 5, x3 + 34, x4 + 19}
          even!_A = 64
          timesIter#!_A(x1,x2,x3) = max{x2 + 31, x3 + 4}
          half!_A(x1) = max{10, x1 - 17}
          s!_A(x1) = 33
          check!_A(x1) = 33
          plus!_A(x1,x2) = max{67, x2}
          odd!_A = 1
          p!_A(x1) = max{5, x1 - 60}
          |0|!_A = 4
          zero!_A = 1
  
    partial status:
  
      pi(if#) = []
      pi(even) = []
      pi(timesIter#) = []
      pi(half) = []
      pi(s) = []
      pi(check) = []
      pi(plus) = []
      pi(odd) = []
      pi(p) = []
      pi(|0|) = []
      pi(zero) = []

The following rules are removed:

  p1, p2, p3, p4

Then no dependency pair remains.

-- Reduction pair.

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

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

and R consists of:

r1: check(|0|()) -> zero()
r2: check(s(|0|())) -> odd()
r3: check(s(s(|0|()))) -> even()
r4: check(s(s(s(x)))) -> check(s(x))
r5: half(|0|()) -> |0|()
r6: half(s(|0|())) -> |0|()
r7: half(s(s(x))) -> s(half(x))
r8: plus(|0|(),y) -> y
r9: plus(s(x),y) -> s(plus(x,y))
r10: times(x,y) -> timesIter(x,y,|0|())
r11: timesIter(x,y,z) -> if(check(x),x,y,z,plus(z,y))
r12: p(s(x)) -> x
r13: p(|0|()) -> |0|()
r14: if(zero(),x,y,z,u) -> z
r15: if(odd(),x,y,z,u) -> timesIter(p(x),y,u)
r16: if(even(),x,y,z,u) -> plus(timesIter(half(x),y,half(z)),timesIter(half(x),y,half(s(z))))

The set of usable rules consists of

  r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        check#_A(x1) = max{18, x1}
        s_A(x1) = x1 + 6
        check_A(x1) = max{4, x1 - 19}
        |0|_A = 2
        odd_A = 1
        even_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          check#_A(x1) = 0
          s_A(x1) = max{179, x1 - 8}
          check_A(x1) = x1 + 94
          |0|_A = 84
          odd_A = 171
          even_A = 1
          check#!_A(x1) = x1 + 26
          s!_A(x1) = max{215, x1 + 35}
          check!_A(x1) = max{215, x1 - 67}
          |0|!_A = 114
          odd!_A = 8
          even!_A = 113
  
    partial status:
  
      pi(check#) = [1]
      pi(s) = [1]
      pi(check) = []
      pi(|0|) = []
      pi(odd) = []
      pi(even) = []

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

and R consists of:

r1: check(|0|()) -> zero()
r2: check(s(|0|())) -> odd()
r3: check(s(s(|0|()))) -> even()
r4: check(s(s(s(x)))) -> check(s(x))
r5: half(|0|()) -> |0|()
r6: half(s(|0|())) -> |0|()
r7: half(s(s(x))) -> s(half(x))
r8: plus(|0|(),y) -> y
r9: plus(s(x),y) -> s(plus(x,y))
r10: times(x,y) -> timesIter(x,y,|0|())
r11: timesIter(x,y,z) -> if(check(x),x,y,z,plus(z,y))
r12: p(s(x)) -> x
r13: p(|0|()) -> |0|()
r14: if(zero(),x,y,z,u) -> z
r15: if(odd(),x,y,z,u) -> timesIter(p(x),y,u)
r16: if(even(),x,y,z,u) -> plus(timesIter(half(x),y,half(z)),timesIter(half(x),y,half(s(z))))

The set of usable rules consists of

  r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        half#_A(x1) = 1
        s_A(x1) = 0
        check_A(x1) = max{0, x1 - 6}
        |0|_A = 16
        even_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          half#_A(x1) = 0
          s_A(x1) = max{42, x1 + 3}
          check_A(x1) = max{52, x1 - 45}
          |0|_A = 14
          even_A = 1
          half#!_A(x1) = max{4, x1 - 1}
          s!_A(x1) = max{3, x1 - 80}
          check!_A(x1) = max{2, x1 - 88}
          |0|!_A = 5
          even!_A = 2
  
    partial status:
  
      pi(half#) = []
      pi(s) = []
      pi(check) = []
      pi(|0|) = []
      pi(even) = []

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: plus#(s(x),y) -> plus#(x,y)

and R consists of:

r1: check(|0|()) -> zero()
r2: check(s(|0|())) -> odd()
r3: check(s(s(|0|()))) -> even()
r4: check(s(s(s(x)))) -> check(s(x))
r5: half(|0|()) -> |0|()
r6: half(s(|0|())) -> |0|()
r7: half(s(s(x))) -> s(half(x))
r8: plus(|0|(),y) -> y
r9: plus(s(x),y) -> s(plus(x,y))
r10: times(x,y) -> timesIter(x,y,|0|())
r11: timesIter(x,y,z) -> if(check(x),x,y,z,plus(z,y))
r12: p(s(x)) -> x
r13: p(|0|()) -> |0|()
r14: if(zero(),x,y,z,u) -> z
r15: if(odd(),x,y,z,u) -> timesIter(p(x),y,u)
r16: if(even(),x,y,z,u) -> plus(timesIter(half(x),y,half(z)),timesIter(half(x),y,half(s(z))))

The set of usable rules consists of

  r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = x1 + 1
        s_A(x1) = x1 + 26
        check_A(x1) = max{8, x1 - 25}
        |0|_A = 21
        odd_A = 11
        even_A = 9
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = 0
          s_A(x1) = max{27, x1 - 17}
          check_A(x1) = max{0, x1 - 1}
          |0|_A = 12
          odd_A = 3
          even_A = 1
          plus#!_A(x1,x2) = max{8, x2 + 1}
          s!_A(x1) = max{7, x1 - 40}
          check!_A(x1) = max{2, x1 - 58}
          |0|!_A = 4
          odd!_A = 1
          even!_A = 1
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = []
      pi(check) = []
      pi(|0|) = []
      pi(odd) = []
      pi(even) = []

The following rules are removed:

  p1

Then no dependency pair remains.