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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if#_A(x1,x2,x3,x4,x5) = max{0, x1 - 2, x2 - 1}
        even_A = 119
        timesIter#_A(x1,x2,x3) = x1
        half_A(x1) = max{54, x1 - 49}
        s_A(x1) = max{152, x1 + 50}
        check_A(x1) = x1 + 1
        plus_A(x1,x2) = max{6, x1 - 43}
        odd_A = 21
        p_A(x1) = max{18, x1 - 20}
        |0|_A = 17
        zero_A = 1
  
    base order 2:
  
      precedence:
      
        plus > even = s > timesIter# > if# = check > half = odd > p > |0| > zero
      
  
    partial status:
  
      pi(if#) = []
      pi(even) = []
      pi(timesIter#) = [1]
      pi(half) = []
      pi(s) = []
      pi(check) = [1]
      pi(plus) = []
      pi(odd) = []
      pi(p) = []
      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

  r1, r2, r3, r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        check#_A(x1) = max{13, x1 - 31}
        s_A(x1) = max{56, x1 + 15}
        check_A(x1) = max{11, x1 - 58}
        |0|_A = 13
        zero_A = 0
        odd_A = 10
        even_A = 12
        half_A(x1) = max{18, x1 - 32}
  
    base order 2:
  
      precedence:
      
        s > half > check# > |0| > check > zero = odd = even
      
  
    partial status:
  
      pi(check#) = []
      pi(s) = [1]
      pi(check) = []
      pi(|0|) = []
      pi(zero) = []
      pi(odd) = []
      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

  r1, r2, r3, r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        half#_A(x1) = x1 + 7
        s_A(x1) = x1 + 8
        check_A(x1) = max{2, x1 - 25}
        |0|_A = 6
        zero_A = 0
        odd_A = 1
        even_A = 1
        half_A(x1) = max{7, x1 - 17}
  
    base order 2:
  
      precedence:
      
        |0| > s > half# = half > check > zero = odd = even
      
  
    partial status:
  
      pi(half#) = [1]
      pi(s) = [1]
      pi(check) = []
      pi(|0|) = []
      pi(zero) = []
      pi(odd) = []
      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: 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

  r1, r2, r3, r5, r6, r13

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = x1
        s_A(x1) = x1
        check_A(x1) = max{10, x1 - 9}
        |0|_A = 0
        zero_A = 0
        odd_A = 0
        even_A = 0
        half_A(x1) = max{0, x1 - 2}
        p_A(x1) = max{7, x1 - 8}
  
    base order 2:
  
      precedence:
      
        s > plus# = half > check = p > |0| > zero = odd = even
      
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(check) = []
      pi(|0|) = []
      pi(zero) = []
      pi(odd) = []
      pi(even) = []
      pi(half) = []
      pi(p) = []

The following rules are removed:

  p1

Then no dependency pair remains.