YES

We show the termination of the TRS R:

  minus(minus(x)) -> x
  minus(h(x)) -> h(minus(x))
  minus(f(x,y)) -> f(minus(y),minus(x))

-- SCC decomposition.

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

p1: minus#(h(x)) -> minus#(x)
p2: minus#(f(x,y)) -> minus#(y)
p3: minus#(f(x,y)) -> minus#(x)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(h(x)) -> h(minus(x))
r3: minus(f(x,y)) -> f(minus(y),minus(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: minus#(h(x)) -> minus#(x)
p2: minus#(f(x,y)) -> minus#(x)
p3: minus#(f(x,y)) -> minus#(y)

and R consists of:

r1: minus(minus(x)) -> x
r2: minus(h(x)) -> h(minus(x))
r3: minus(f(x,y)) -> f(minus(y),minus(x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        minus# > f > h
      
      argument filter:
    
        pi(minus#) = 1
        pi(h) = [1]
        pi(f) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        minus# > f > h
      
      argument filter:
    
        pi(minus#) = [1]
        pi(h) = 1
        pi(f) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        minus# > f > h
      
      argument filter:
    
        pi(minus#) = [1]
        pi(h) = [1]
        pi(f) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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