YES

We show the termination of the TRS R:

  double(|0|()) -> |0|()
  double(s(x)) -> s(s(double(x)))
  half(|0|()) -> |0|()
  half(s(|0|())) -> |0|()
  half(s(s(x))) -> s(half(x))
  -(x,|0|()) -> x
  -(s(x),s(y)) -> -(x,y)
  if(|0|(),y,z) -> y
  if(s(x),y,z) -> z
  half(double(x)) -> x

-- SCC decomposition.

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

p1: double#(s(x)) -> double#(x)
p2: half#(s(s(x))) -> half#(x)
p3: -#(s(x),s(y)) -> -#(x,y)

and R consists of:

r1: double(|0|()) -> |0|()
r2: double(s(x)) -> s(s(double(x)))
r3: half(|0|()) -> |0|()
r4: half(s(|0|())) -> |0|()
r5: half(s(s(x))) -> s(half(x))
r6: -(x,|0|()) -> x
r7: -(s(x),s(y)) -> -(x,y)
r8: if(|0|(),y,z) -> y
r9: if(s(x),y,z) -> z
r10: half(double(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2}
  {p3}


-- Reduction pair.

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

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

and R consists of:

r1: double(|0|()) -> |0|()
r2: double(s(x)) -> s(s(double(x)))
r3: half(|0|()) -> |0|()
r4: half(s(|0|())) -> |0|()
r5: half(s(s(x))) -> s(half(x))
r6: -(x,|0|()) -> x
r7: -(s(x),s(y)) -> -(x,y)
r8: if(|0|(),y,z) -> y
r9: if(s(x),y,z) -> z
r10: half(double(x)) -> x

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      double#_A(x1) = ((1,0),(1,0)) x1
      s_A(x1) = ((1,1),(1,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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: double(|0|()) -> |0|()
r2: double(s(x)) -> s(s(double(x)))
r3: half(|0|()) -> |0|()
r4: half(s(|0|())) -> |0|()
r5: half(s(s(x))) -> s(half(x))
r6: -(x,|0|()) -> x
r7: -(s(x),s(y)) -> -(x,y)
r8: if(|0|(),y,z) -> y
r9: if(s(x),y,z) -> z
r10: half(double(x)) -> x

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      half#_A(x1) = ((1,1),(0,0)) x1
      s_A(x1) = ((1,1),(0,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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: -#(s(x),s(y)) -> -#(x,y)

and R consists of:

r1: double(|0|()) -> |0|()
r2: double(s(x)) -> s(s(double(x)))
r3: half(|0|()) -> |0|()
r4: half(s(|0|())) -> |0|()
r5: half(s(s(x))) -> s(half(x))
r6: -(x,|0|()) -> x
r7: -(s(x),s(y)) -> -(x,y)
r8: if(|0|(),y,z) -> y
r9: if(s(x),y,z) -> z
r10: half(double(x)) -> x

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      -#_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(0,1)) x2
      s_A(x1) = ((1,1),(1,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1

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