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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          if#_A(x1,x2,x3,x4,x5) = 1
          even_A() = 12
          timesIter#_A(x1,x2,x3) = 1
          half_A(x1) = x1 + 2
          s_A(x1) = x1 + 3
          check_A(x1) = x1 + 5
          plus_A(x1,x2) = x1 + 5
          odd_A() = 3
          p_A(x1) = x1 + 2
          |0|_A() = 13
          zero_A() = 12
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if#_A(x1,x2,x3,x4,x5) = max{21, x1 + 1, x2 - 3}
          even_A = 19
          timesIter#_A(x1,x2,x3) = max{21, x1 - 1}
          half_A(x1) = max{7, x1 - 7}
          s_A(x1) = max{14, x1 + 10}
          check_A(x1) = max{0, x1 - 6}
          plus_A(x1,x2) = max{11, x1 - 15}
          odd_A = 9
          p_A(x1) = max{5, x1 - 2}
          |0|_A = 5
          zero_A = 0
          if#!_A(x1,x2,x3,x4,x5) = max{0, x1 - 7, x2 - 7}
          even!_A = 15
          timesIter#!_A(x1,x2,x3) = max{1, x1 - 6}
          half!_A(x1) = max{16, x1 - 101}
          s!_A(x1) = 13
          check!_A(x1) = 17
          plus!_A(x1,x2) = max{14, x1 - 25, x2 - 2}
          odd!_A = 19
          p!_A(x1) = 0
          |0|!_A = 18
          zero!_A = 17
  
    partial status:
  
      pi(if#) = []
      pi(even) = []
      pi(timesIter#) = []
      pi(half) = []
      pi(s) = []
      pi(check) = [1]
      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

  r3, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          check#_A(x1) = 6
          s_A(x1) = 7
          check_A(x1) = x1 + 3
          |0|_A() = 4
          even_A() = 3
          half_A(x1) = 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          check#_A(x1) = x1
          s_A(x1) = x1 + 48
          check_A(x1) = max{28, x1 - 152}
          |0|_A = 56
          even_A = 1
          half_A(x1) = max{26, x1 - 34}
          check#!_A(x1) = max{51, x1 + 21}
          s!_A(x1) = max{0, x1 - 28}
          check!_A(x1) = 3
          |0|!_A = 6
          even!_A = 2
          half!_A(x1) = max{5, x1 - 97}
  
    partial status:
  
      pi(check#) = []
      pi(s) = []
      pi(check) = [1]
      pi(|0|) = []
      pi(even) = []
      pi(half) = []

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

  r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          half#_A(x1) = x1
          s_A(x1) = x1 + 7
          check_A(x1) = 6
          |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:
        
          half#_A(x1) = 0
          s_A(x1) = max{22, x1 - 14}
          check_A(x1) = max{26, x1 - 30}
          |0|_A = 29
          odd_A = 1
          even_A = 1
          half#!_A(x1) = max{10, x1 - 40}
          s!_A(x1) = max{11, x1 - 27}
          check!_A(x1) = max{8, x1 - 42}
          |0|!_A = 10
          odd!_A = 3
          even!_A = 1
  
    partial status:
  
      pi(half#) = []
      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: 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

  r3, r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x1 + x2 + 16
          s_A(x1) = x1 + 15
          check_A(x1) = x1 + 6
          |0|_A() = 2
          even_A() = 31
          half_A(x1) = x1 + 5
  
    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{54, x1 - 28}
          check_A(x1) = max{61, x1 - 55}
          |0|_A = 4
          even_A = 1
          half_A(x1) = max{27, x1 - 56}
          plus#!_A(x1,x2) = 1
          s!_A(x1) = 5
          check!_A(x1) = max{1, x1 - 47}
          |0|!_A = 4
          even!_A = 3
          half!_A(x1) = max{6, x1 - 19}
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(check) = [1]
      pi(|0|) = []
      pi(even) = []
      pi(half) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.