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:
  
      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:
          
            if#_A(x1,x2,x3,x4,x5) = max{1, x2 - 4, x3 - 2}
            even_A = 62
            timesIter#_A(x1,x2,x3) = max{1, x1 - 4, x2 - 2}
            half_A(x1) = max{3, x1 - 28}
            s_A(x1) = max{56, x1 + 30}
            check_A(x1) = max{45, x1 + 1}
            plus_A(x1,x2) = max{109, x1 - 57}
            odd_A = 57
            p_A(x1) = max{5, x1 - 30}
            |0|_A = 0
            zero_A = 2
            if#!_A(x1,x2,x3,x4,x5) = max{x1 - 46, x2 + 4}
            even!_A = 24
            timesIter#!_A(x1,x2,x3) = x1 + 5
            half!_A(x1) = max{16, x1 - 31}
            s!_A(x1) = 18
            check!_A(x1) = max{23, x1 - 61}
            plus!_A(x1,x2) = max{23, x1 - 57, x2 - 2}
            odd!_A = 2
            p!_A(x1) = max{3, x1 - 1}
            |0|!_A = 1
            zero!_A = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if#_A(x1,x2,x3,x4,x5) = 0
              even_A() = 1
              timesIter#_A(x1,x2,x3) = 0
              half_A(x1) = x1
              s_A(x1) = x1 + 8
              check_A(x1) = x1 + 1
              plus_A(x1,x2) = x1 + 1
              odd_A() = 10
              p_A(x1) = x1 + 1
              |0|_A() = 1
              zero_A() = 1
              if#!_A(x1,x2,x3,x4,x5) = x2 + x3 + x4 + 17
              even!_A() = 14
              timesIter#!_A(x1,x2,x3) = 13
              half!_A(x1) = x1 + 1
              s!_A(x1) = 16
              check!_A(x1) = 15
              plus!_A(x1,x2) = x1 + x2 + 17
              odd!_A() = 12
              p!_A(x1) = 13
              |0|!_A() = 11
              zero!_A() = 0
        
  
    partial status:
  
      pi(if#) = []
      pi(even) = []
      pi(timesIter#) = []
      pi(half) = []
      pi(s) = [1]
      pi(check) = []
      pi(plus) = []
      pi(odd) = []
      pi(p) = [1]
      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

  (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:
          
            check#_A(x1) = x1
            s_A(x1) = 8
            check#!_A(x1) = max{21, x1 - 9}
            s!_A(x1) = 21
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              check#_A(x1) = x1
              s_A(x1) = x1 + 3
              check#!_A(x1) = x1
              s!_A(x1) = x1 + 2
        
  
    partial status:
  
      pi(check#) = []
      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: 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

  (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:
          
            half#_A(x1) = 0
            s_A(x1) = 0
            half#!_A(x1) = 1
            s!_A(x1) = 43
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              half#_A(x1) = x1
              s_A(x1) = x1 + 2
              half#!_A(x1) = x1 + 6
              s!_A(x1) = x1 + 7
        
  
    partial status:
  
      pi(half#) = []
      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: 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

  (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:
          
            plus#_A(x1,x2) = x1
            s_A(x1) = max{47, x1 + 16}
            plus#!_A(x1,x2) = max{5, x1 - 17}
            s!_A(x1) = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = 0
              s_A(x1) = 2
              plus#!_A(x1,x2) = 3
              s!_A(x1) = 2
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.