YES

We show the termination of the TRS R:

  +(|0|(),y) -> y
  +(s(x),y) -> s(+(x,y))
  -(|0|(),y) -> |0|()
  -(x,|0|()) -> x
  -(s(x),s(y)) -> -(x,y)

-- SCC decomposition.

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

p1: +#(s(x),y) -> +#(x,y)
p2: -#(s(x),s(y)) -> -#(x,y)

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: -(|0|(),y) -> |0|()
r4: -(x,|0|()) -> x
r5: -(s(x),s(y)) -> -(x,y)

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2}


-- Reduction pair.

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

p1: +#(s(x),y) -> +#(x,y)

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: -(|0|(),y) -> |0|()
r4: -(x,|0|()) -> x
r5: -(s(x),s(y)) -> -(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        +#_A(x1,x2) = max{0, x1 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      +# = s
    
    partial status:
  
      pi(+#) = []
      pi(s) = [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: -#(s(x),s(y)) -> -#(x,y)

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: -(|0|(),y) -> |0|()
r4: -(x,|0|()) -> x
r5: -(s(x),s(y)) -> -(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        -#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      -# = s
    
    partial status:
  
      pi(-#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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