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!):
        
          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 + 3
              check_A(x1) = 1
              plus_A(x1,x2) = 0
              odd_A() = 1
              p_A(x1) = x1
              |0|_A() = 8
              zero_A() = 0
              if#!_A(x1,x2,x3,x4,x5) = x1 + x2 + 3
              even!_A() = 6
              timesIter#!_A(x1,x2,x3) = x1 + 4
              half!_A(x1) = x1
              s!_A(x1) = 5
              check!_A(x1) = x1 + 5
              plus!_A(x1,x2) = 10
              odd!_A() = 6
              p!_A(x1) = 4
              |0|!_A() = 7
              zero!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if#_A(x1,x2,x3,x4,x5) = max{1, x2 - 38, x3 - 2}
            even_A = 124
            timesIter#_A(x1,x2,x3) = max{1, x1 - 38, x2 - 2}
            half_A(x1) = max{37, x1 - 17}
            s_A(x1) = max{72, x1 + 54}
            check_A(x1) = max{2, x1 - 2}
            plus_A(x1,x2) = max{21, x1 - 55}
            odd_A = 56
            p_A(x1) = max{39, x1 - 18}
            |0|_A = 3
            zero_A = 2
            if#!_A(x1,x2,x3,x4,x5) = max{2, x1 - 41, x2 - 55}
            even!_A = 31
            timesIter#!_A(x1,x2,x3) = max{15, x1 - 38}
            half!_A(x1) = 84
            s!_A(x1) = max{82, x1 - 25}
            check!_A(x1) = max{30, x1 - 2}
            plus!_A(x1,x2) = max{83, x1 - 3, x2 - 3}
            odd!_A = 71
            p!_A(x1) = max{56, x1 - 4}
            |0|!_A = 81
            zero!_A = 82
        
  
    partial status:
  
      pi(if#) = []
      pi(even) = []
      pi(timesIter#) = []
      pi(half) = []
      pi(s) = [1]
      pi(check) = [1]
      pi(plus) = []
      pi(odd) = []
      pi(p) = [1]
      pi(|0|) = []
      pi(zero) = []

The following rules are removed:

  p2



-- SCC decomposition.

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: if#(even(),x,y,z,u) -> timesIter#(half(x),y,half(z))
p3: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              check#_A(x1) = x1
              s_A(x1) = 2
              check#!_A(x1) = 0
              s!_A(x1) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            check#_A(x1) = 0
            s_A(x1) = max{45, x1 - 4}
            check#!_A(x1) = max{1, x1 - 46}
            s!_A(x1) = 2
        
  
    partial status:
  
      pi(check#) = [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: 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

  r5, r6

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              half#_A(x1) = x1
              s_A(x1) = x1 + 5
              half_A(x1) = x1 + 4
              |0|_A() = 2
              half#!_A(x1) = x1 + 1
              s!_A(x1) = x1 + 2
              half!_A(x1) = 5
              |0|!_A() = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            half#_A(x1) = 0
            s_A(x1) = 0
            half_A(x1) = max{23, x1 - 18}
            |0|_A = 0
            half#!_A(x1) = 1
            s!_A(x1) = max{61, x1 + 8}
            half!_A(x1) = max{30, x1 - 29}
            |0|!_A = 34
        
  
    partial status:
  
      pi(half#) = []
      pi(s) = []
      pi(half) = []
      pi(|0|) = []

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = x1
              s_A(x1) = x1 + 3
              plus#!_A(x1,x2) = x1 + 1
              s!_A(x1) = x1 + 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = x1
            s_A(x1) = max{75, x1 + 5}
            plus#!_A(x1,x2) = max{7, x1 - 76}
            s!_A(x1) = max{7, x1 - 41}
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.