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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      minus#_A(x1) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(1,1,0,0)) x1
      h_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x1 + (1,1,1,1)
      f_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,0,0),(1,1,1,0)) x2 + (1,1,1,1)

The next rules are strictly ordered:

  p1, p2, p3

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