YES

We show the termination of the TRS R:

  even(|0|()) -> true()
  even(s(|0|())) -> false()
  even(s(s(x))) -> even(x)
  half(|0|()) -> |0|()
  half(s(s(x))) -> s(half(x))
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))
  times(|0|(),y) -> |0|()
  times(s(x),y) -> if_times(even(s(x)),s(x),y)
  if_times(true(),s(x),y) -> plus(times(half(s(x)),y),times(half(s(x)),y))
  if_times(false(),s(x),y) -> plus(y,times(x,y))

-- SCC decomposition.

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

p1: even#(s(s(x))) -> even#(x)
p2: half#(s(s(x))) -> half#(x)
p3: plus#(s(x),y) -> plus#(x,y)
p4: times#(s(x),y) -> if_times#(even(s(x)),s(x),y)
p5: times#(s(x),y) -> even#(s(x))
p6: if_times#(true(),s(x),y) -> plus#(times(half(s(x)),y),times(half(s(x)),y))
p7: if_times#(true(),s(x),y) -> times#(half(s(x)),y)
p8: if_times#(true(),s(x),y) -> half#(s(x))
p9: if_times#(false(),s(x),y) -> plus#(y,times(x,y))
p10: if_times#(false(),s(x),y) -> times#(x,y)

and R consists of:

r1: even(|0|()) -> true()
r2: even(s(|0|())) -> false()
r3: even(s(s(x))) -> even(x)
r4: half(|0|()) -> |0|()
r5: half(s(s(x))) -> s(half(x))
r6: plus(|0|(),y) -> y
r7: plus(s(x),y) -> s(plus(x,y))
r8: times(|0|(),y) -> |0|()
r9: times(s(x),y) -> if_times(even(s(x)),s(x),y)
r10: if_times(true(),s(x),y) -> plus(times(half(s(x)),y),times(half(s(x)),y))
r11: if_times(false(),s(x),y) -> plus(y,times(x,y))

The estimated dependency graph contains the following SCCs:

  {p4, p7, p10}
  {p1}
  {p2}
  {p3}


-- Reduction pair.

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

p1: if_times#(false(),s(x),y) -> times#(x,y)
p2: times#(s(x),y) -> if_times#(even(s(x)),s(x),y)
p3: if_times#(true(),s(x),y) -> times#(half(s(x)),y)

and R consists of:

r1: even(|0|()) -> true()
r2: even(s(|0|())) -> false()
r3: even(s(s(x))) -> even(x)
r4: half(|0|()) -> |0|()
r5: half(s(s(x))) -> s(half(x))
r6: plus(|0|(),y) -> y
r7: plus(s(x),y) -> s(plus(x,y))
r8: times(|0|(),y) -> |0|()
r9: times(s(x),y) -> if_times(even(s(x)),s(x),y)
r10: if_times(true(),s(x),y) -> plus(times(half(s(x)),y),times(half(s(x)),y))
r11: if_times(false(),s(x),y) -> plus(y,times(x,y))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        if_times#_A(x1,x2,x3) = ((0,1),(0,0)) x1 + x2
        false_A() = (2,0)
        s_A(x1) = x1 + (2,0)
        times#_A(x1,x2) = x1 + (1,0)
        even_A(x1) = x1 + (0,1)
        true_A() = (0,1)
        half_A(x1) = x1 + (0,1)
        |0|_A() = (0,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        if_times#_A(x1,x2,x3) = x2 + (3,0)
        false_A() = (1,1)
        s_A(x1) = ((0,1),(0,0)) x1 + (1,4)
        times#_A(x1,x2) = ((1,1),(0,0)) x1 + (0,4)
        even_A(x1) = (2,1)
        true_A() = (1,1)
        half_A(x1) = ((0,0),(1,0)) x1 + (2,0)
        |0|_A() = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        if_times#_A(x1,x2,x3) = ((1,1),(1,1)) x2
        false_A() = (3,2)
        s_A(x1) = (2,1)
        times#_A(x1,x2) = ((1,1),(0,0)) x1 + (1,3)
        even_A(x1) = (2,1)
        true_A() = (1,1)
        half_A(x1) = (0,1)
        |0|_A() = (0,1)
    

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

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

and R consists of:

r1: even(|0|()) -> true()
r2: even(s(|0|())) -> false()
r3: even(s(s(x))) -> even(x)
r4: half(|0|()) -> |0|()
r5: half(s(s(x))) -> s(half(x))
r6: plus(|0|(),y) -> y
r7: plus(s(x),y) -> s(plus(x,y))
r8: times(|0|(),y) -> |0|()
r9: times(s(x),y) -> if_times(even(s(x)),s(x),y)
r10: if_times(true(),s(x),y) -> plus(times(half(s(x)),y),times(half(s(x)),y))
r11: if_times(false(),s(x),y) -> plus(y,times(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        even#_A(x1) = ((1,0),(1,0)) x1
        s_A(x1) = x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        even#_A(x1) = ((1,0),(1,0)) x1
        s_A(x1) = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        even#_A(x1) = (0,0)
        s_A(x1) = (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  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: even(|0|()) -> true()
r2: even(s(|0|())) -> false()
r3: even(s(s(x))) -> even(x)
r4: half(|0|()) -> |0|()
r5: half(s(s(x))) -> s(half(x))
r6: plus(|0|(),y) -> y
r7: plus(s(x),y) -> s(plus(x,y))
r8: times(|0|(),y) -> |0|()
r9: times(s(x),y) -> if_times(even(s(x)),s(x),y)
r10: if_times(true(),s(x),y) -> plus(times(half(s(x)),y),times(half(s(x)),y))
r11: if_times(false(),s(x),y) -> plus(y,times(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        half#_A(x1) = x1
        s_A(x1) = x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        half#_A(x1) = ((0,1),(0,0)) x1
        s_A(x1) = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        half#_A(x1) = (0,0)
        s_A(x1) = (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  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: even(|0|()) -> true()
r2: even(s(|0|())) -> false()
r3: even(s(s(x))) -> even(x)
r4: half(|0|()) -> |0|()
r5: half(s(s(x))) -> s(half(x))
r6: plus(|0|(),y) -> y
r7: plus(s(x),y) -> s(plus(x,y))
r8: times(|0|(),y) -> |0|()
r9: times(s(x),y) -> if_times(even(s(x)),s(x),y)
r10: if_times(true(),s(x),y) -> plus(times(half(s(x)),y),times(half(s(x)),y))
r11: if_times(false(),s(x),y) -> plus(y,times(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,1),(1,1)) x1
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,1),(0,1)) x1
        s_A(x1) = ((1,0),(1,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((0,0),(1,0)) x1
        s_A(x1) = ((1,1),(0,0)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.